IN  MEMORIAM 
FLOR1AN  CAJORI 


BOWSER'S  MATHEMATICS. 


ACADEMIC  ALGEBRA.    With  numerous  Examples. 
COLLEGE  ALGEBRA.     With  numerous  Examples. 

PLANE  AND  SOLID  GEOMETRY.  With  numerous  Exer- 
cises. 

ELEMENTS  OF  PLANE  AND  SPHERICAL  TRIGONOME- 
TRY.  With  numerous  Examples. 

A  TREATISE  ON  PLANE  AND  SPHERICAL  TRIGONOME- 
TRY, and  its  applications  to  Astronomy  and  Geodesy. 

With  numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC  GEOMETRY, 

embracing  Plane  Geometry,  and  an  Introduction  to 
Geometry  of  Three   Dimensions. 

AN  ELEMENTARY  TREATISE  ON  THE  DIFFERENTIAL 
AND  INTEGRAL  CALCULUS.  With  numerous  Exam- 
plea. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC  MECHANICS. 

With  numerous  Examples. 

AN    ELEMENTARY     TREATISE     ON    HYDROMECHANICS. 

With  numerous  Examples. 


A   TREATISE 

ON 

PLANE   AND   SPHERICAL 

TRIGONOMETRY, 


AND   ITS   APPLICATIONS   TO 


ASTBONOMY  AND   GEODESY, 


NUMEROUS    EXAMPLES. 


BY 

EDWARD   A.   BOWSER,   LL.D., 

PROFESSOR  OF  MATHEMATICS  AND  ENGINEERING  IN  RUTGERS  COLLEGE. 


BOSTON,  U.S.A. : 
PUBLISHED  BY   D.  C.  HEATH   &   CO. 

1892. 


COPYRIGHT,  1892, 
BY  E.  A.  BOWSER. 


CAJORI 


TYPOGRAPHY  BY  J.  S.  GUSHING  &  Co.,  BOSTON,  U.S.A. 
PRESSWORK  BY  BERWICK  &  SMITH,  BOSTON,  U.S.A. 


PEEFACB. 


THE  present  treatise  on  Plane  and  Spherical  Trigo- 
nometry is  designed  as  a  text-book  for  Colleges,  Scien- 
tific Schools,  and  Institutes  of  Technology.  The  aim 
has  been  to  present  the  subject  in  as  concise  a  form  as 
is  consistent  with  clearness,  to  make  it  attractive  and 
easily  intelligible  to  the  student,  and  at  the  same 
time  to  present  the  fullest  course  of  Trigonometry 
which  is  usually  given  in  the  best  Technological 
Schools. 

Considerable  care  has  been  taken  to  instruct  the 
student  in  the  theory  and  use  of  Logarithms,  and 
their  practical  application  to  the  solution  of  triangles. 
It  is  hoped  that  the  work  may  commend  itself,  not 
only  to  those  who  wish  to  confine  themselves  to  the 
numerical  calculations  which  occur  in  Trigonometry, 
but  also  to  those  who  intend  to  pursue  the  study  of 
the  higher  mathematics. 

The  examples  are  very  numerous  and  are  carefully 
selected.  Many  are  placed  in  immediate  connection 
with  the  subject-matter  which  they  illustrate.  The 
numerical  solution  of  triangles  has  received  much 
attention,  each  case  being  treated  in  detail.  The 

911384 


IV  PREFACE. 

examples  at  the  ends  of  the  chapters  have  been  care- 
fully graded,  beginning  with  those  which  are  easy, 
and  extending  to  those  which  are  more  and  more  diffi- 
cult. These  examples  illustrate  every  part  of  the  sub- 
ject, and  are  intended  to  test,  not  only  the  student's 
knowledge  of  the  usual  methods  of  computation,  but 
his  ability  to  grasp  them  in  the  many  forms  they  may 
assume  in  practical  applications.  Among  these  exam- 
ples are  some  of  the  most  elegant  theorems  in  Plane 
and  Spherical  Trigonometry. 

The  Chapters  on  De  Moivre's  Theorem,  and  Astron- 
omy, Geodesy,  and  Polyedrons,  will  serve  to  introduce 
the  student  to  some  of  the  higher  applications  of 
Trigonometry,  rarely  found  in  American  text-books. 

In  writing  this  book,  the  best  English  and  French 
authors  have  been  consulted.  I  am  indebted  especially 
to  the  works  of  Todhunter,  Casey,  Lock,  Hobson, 
Clarke,  Eustis,  Snowball,  M'Clelland  and  Preston, 
Smith,  and  Serret. 

It  remains  for  me  to  express  my  thanks  to  my  col- 
leagues, Prof.  R.  W.  Prentiss  for  reading  the  MS., 
and  Mr.  I.  S.  Upson  for  reading  the  proof-sheets. 

Any  corrections  or  suggestions,  either  in  the  text 
or  the  examples,  will  be  thankfully  received. 

E.  A.  B. 
RUTGERS  COLL,KOK, 

New  Brunswick,  N.  J.,  April,  181L'. 


TABLE     OF     CONTENTS. 

PART  I. 
PLANE   TRIGONOMETRY. 

CHAPTER   I. 

MEASUREMENT  or  ANGLES. 

RT.  PAGE 

1.  Trigonometry 1 

2.  The  Measure  of  a  Quantity 1 

3.  Angles 2 

4.  Positive  and  Negative  Angles 3 

5.  The  Measure  of  Angles 3 

0.  The  Sexagesimal  Method 5 

7.  The  Centesimal  or  Decimal  Method 6 

8.  The  Circular  Measure 6 

9.  Comparison  of  the  Sexagesimal  and  Centesimal  Measures  . .  8 

10.  Comparison  of  the  Sexagesimal  and  Circular  Measures 9 

11.  General  Measure  of  an  Angle 11 

12.  Complement  and  Supplement  of  an  Angle 12 

Examples 13 

CHAPTER   II. 
THE  TRIGONOMETRIC  FUNCTIONS. 

13.  Definitions  of  the  Trigonometric  Functions 16 

14.  The  Functions  are  always  the  Same  for  the  Same  Angle 18 

15.  Functions  of  Complemental  Angles 20 

16.  Representation  of  the  Functions  by  Straight  Lines 20 

17.  Positive  and  Negative  Lines 23 

v 


yi  CON TENTS. 

ART.  PAGE 

18.  Functions  of  Angles  of  Any  Magnitude 23 

19.  Changes  in  Sine  as  the  Angle  increases  from  0°  to  360° 25 

20.  Changes  in  Cosine  as  the  Angle  increases  from  0°  to  360°. . .  26 

21.  Changes  in  Tangent  as  the  Angle  increases  from  0°  to  360°.  27 

22.  Table  giving  Changes  of  Functions  in  Four  Quadrants 28 

23.  Relations  between  the  Functions  of  the  Same  Angle 29 

24.  Use  of  the  Preceding  Formulae 30 

25.  Graphic  Method  of  finding  the  Functions  in  Terms  of  One . .  30 

26.  To  find  the  Trigonometric  Functions  of  45° 31 

27.  To  find  the  Trigonometric  Functions  of  60°  and  30° 31 

28.  Reduction  of  Functions  to  1st  Quadrant 33 

29.  Functions  of  Complemental  Angles 34 

30.  Functions  of  Supplemental  Angles 34 

31.  To  prove  sin  (90°  +  A)  =  cos  A,  etc 35 

32.  To  prove  sin  (180°  +  A)  =  -  sin  A,  etc 35 

33.  To  prove  sin  (—  A)  =  —  sin  A,  etc 36 

34.  To  prove  sin  (270°  +  A)  =  sin  (270°  -  A)  =  -  cos  A,  etc 36 

35.  Table  giving  the  Reduced  Functions  of  Any  Angle 37 

36.  Periodicity  of  the  Trigonometric  Functions 38 

37.  Angles  corresponding  to  Given  Functions 39 

38.  General  Expression  for  All  Angles  with  a  Given  Sine 40 

39.  An  Expression  for  All  Angles  with  a  Given  Cosine 41 

40.  An  Expression  for  All  Angles  with  a  Given  Tangent 41 

41.  Trigonometric  Identities 43 

Examples 44 


CHAPTER   III. 
TRIGONOMETRIC  FUNCTIONS  OF  Two  ANGLES. 

42.  Fundamental  Formulae 50 

43.  To  find  the  Values  of  sin  (  x  +  y)  and  cos  (x  +  y) 50 

44.  To  find  the  Values  of  sin  (x  -  y}  and  cos  (x  -  y) 52 

46.  Formulae  for  transforming  Sums  into  Products 55 

46.  Useful  Formulas 56 

47.  Tangent  of  Sum  and  Difference  of  Two  Angles 57 

48.  Formulae  for  the  Sum  of  Three  or  More  Angles 58 

49.  Functions  of  Double  Angles 60 

60.  Functions  of  3  x  in  Terms  of  the  Functions  of  x 61 

61.  Functions  of  Half  an  Angle 63 

62.  Double  Values  of  Sine  and  Cosine  of  Half  an  Angle 63 


CONTENTS.  vii 

ABT.  PAGE 

53.  Quadruple  Values  of  Sine  and  Cosine  of  Half  an  Angle 65 

54.  Double  Value  of  Tangent  of  Half  an  Angle 66 

55.  Triple  Value  of  Sine  of  One-third  an  Angle 67 

56.  Find  the  Values  of  the  Functions  of  22 £° 69 

57.  Find  the  Sine  and  Cosine  of  18° 69 

58.  Find  the  Sine  and  Cosine  of  36° 70 

59.  If  A  +  B  +  C  =  180°,  to  find  sin  A  -f  sin  B  +  sin  C,  etc 70 

60.  Inverse  Trigonometric  Functions 72 

61.  Table  of  Useful  Formulae 75 

Examples  ...   77 


CHAPTER   IV. 
LOGARITHMS  AND  LOGARITHMIC  TABLES.  —  TRIGONOMETRIC  TABLES. 

62.  Nature  and  Use  of  Logarithms 87 

63.  Properties  of  Logarithms 87 

64.  Common  System  of  Logarithms 01 

65.  Comparison  of  Two  Systems  of  Logarithms 93 

66.  Tables  of  Logarithms 95 

67.  Use  of  Tables  of  Logarithms  of  Numbers 98 

68.  To  find  the  Logarithm  of  a  Given  Number 99 

69.  To  find  the  Number  corresponding  to  a  Given  Logarithm . . .   102 
69a.  Arithmetic  Complement 103 

70.  Use  of  Trigonometric  Tables 105 

71.  Use  of  Tables  of  Natural  Trigonometric  Functions 106 

72.  To  find  the  Sine  of  a  Given  Angle 106 

73.  To  find  the  Cosine  of  a  Given  Angle '. 106 

74.  To  find  the  Angle  whose  Sine  is  Given 108 

75.  To  find  the  Angle  whose  Cosine  is  Given 108 

76.  Use  of  Tables  of  Logarithmic  Trigonometric  Functions 110 

77.  To  find  the  Logarithmic  Sine  of  a  Given  Angle 112 

78.  To  find  the  Logarithmic  Cosine  of  a  Given  Angle 112 

79.  To  find  the  Angle  whose  Logarithmic  Sine  is  Given 114 

80.  To  find  the  Angle  whose  Logarithmic  Cosine  is  Given 115 

81.  Angles  near  the  Limits  of  the  Quadrant 116 

Examples 117 


viii  CONTENTS. 

CHAPTER   V. 

SOLUTION  OF  TRIGONOMETRIC  EQUATIONS. 

ART.  PAGE 

82.  Trigonometric  Equations 126 

83.  To  solve  m  sin  0  =  a,  m  cos  <f>  =  b 128 

84.  To  solve  a  sin  0  -f  b  cos  <f>  =  c 129 

85.  To  solve  sin  («  +  x)  =  m  sin  £ 131 

86.  To  solve  tan  («  +  a)  =  m  tan  x 132 

87.  To  solve  tan  («  +  x)  tan  a;  =  m 133 

88.  To  solve  m  sin  (0  +  x)  =  a,  w  sin  (0  +  x}  -  b 134 

89.  To  solve  x  cos  a  -f  y  sin  a  =  w,  £  sin  a  —  y  cos  <t  =  n 135 

90.  Adaptation  to  Logarithmic  Computation 135 

91.  To  solve  r  cos  <£  cos  0  =  a,  r  cos  0  sin  0  =  &,  r  sin  0  =  c 137 

92.  Trigonometric  Elimination 138 

Examples 140 

CHAPTER   VI. 

RELATIONS  BETWEEN  THE  SIDES  OF  A  TRIANGLE  AND  THE  FUNCTIONS 
OF  ITS  ANGLES. 

93.  Formulas 146 

94.  Right  Triangles 146 

95.  Oblique  Triangles  —  Law  of  Sines 147 

96.  Law  of  Cosines 148 

97.  Law  of  Tangents 149 

98.  To  prove  c  =  a  cos  B  +  b  cos  A 149 

99.  Functions  of  Half  an  Angle  in  Terms  of  the  Sides 150 

100.  To  express  the  Sine  of  an  Angle  in  Terms  of  the  Sides 152 

101.  Expressions  for  the  Area  of  a  Triangle 153 

102.  Inscribed  Circle 154 

103.  Circumscribed  Circle 154 

104.  Escribed  Circle 155 

105.  Distance  between  the  In-centre  and  the  Circumcentre 155 

106.  To  find  the  Area  of  a  Cyclic  Quadrilateral 157 

Examples 159 

CHAPTER   VII. 
SOLUTION  OF  TRIANGLES. 

107.  Definitions 165 

108.  Four  Cases  of  Right  Triangles 165 


CONTENTS.  ix 

ART.  PAGE 

109.  Case  I.  —  Given  a  Side  and  the  Hypotenuse 166 

110.  Case  II.  —  Given  an  Acute  Angle  and  the  Hypotenuse 167 

111.  Case  III.  —Given  a  Side  and  an  Acute  Angle 168 

112.  Case  IV.  —  Given  the  Two  Sides 169 

113.  When  a  Side  and  the  Hypotenuse  are  nearly  Equal 169 

114.  Four  Cases  of  Oblique  Triangles 172 

115.  Case  I.  —  Given  a  Side  and  Two  Angles  172 

116.  Case  II. — Given  Two  Sides  and  the  Angle  opposite  One  of  them,  173 

117.  Case  III.  —  Given  Two  Sides  and  the  Included  Angle 176 

118.  Case  IV.  —  Given  the  Three  Sides 177 

119.  Area  of  a  Triangle 180 

120.  Heights  and  Distances  —  Definitions 181 

121.  Heights  of  an  Accessible  Object 182 

122.  Height  and  Distance  of  an  Inaccessible  Object 182 

123.  An  Inaccessible  Object  above  a  Horizontal  Plane 184 

124.  Object  observed  from  Two  Points  in  Same  Vertical  Line ....  185 

125.  Distance  between  Two  Inaccessible  Objects 186 

126.  The  Dip  of  the  Horizon 186 

127.  Problem  of  Pothenot  or  of  Snellius 188 

Examples 189 

CHAPTER   VIII. 
CONSTRUCTION  or  LOGARITHMIC  AND  TRIGONOMETRIC  TABLES. 

128.  Logarithmic  and  Trigonometric  Tables 204 

129.  Exponential  Series 204 

130.  Logarithmic  Series 205 

131.  Computation  of  Logarithms  207 

132.  Sine  and  tan 6  are  in  Ascending  Order  of  Magnitude 208 

133.  The  Limit  of  *^  is  Unity 209 

9 

134.  Limiting  Values  of  sin  6  and  cos  6 209 

135.  To  calculate  the  Sine  and  Cosine  of  10"  and  of  1' 211 

136.  To  construct  a  Table  of  Natural  Sines  and  Cosines 213 

137.  Another  Method 213 

138.  The  Sines  and  Cosines  from  30°  to  60° 214 

139.  Sines  of  Angles  Greater  than  45° 215 

140.  Tables  of  Tangents  and  Secants 215 

141.  Formulae  of  Verification 216 

142.  Tables  of  Logarithmic  Trigonometric  Functions 217 

143.  The  Principle  of  Proportional  Parts 218 


X  CONTENTS. 

ART.  PAGE 

144.  To  prove  the  Rule  for  the  Table  of  Common  Logarithms  ...  218 

145.  To  prove  the  Rule  for  the  Table  of  Natural  Sines 219 

146.  To  prove  the  Rule  for  a  Table  of  Natural  Cosines 219 

147.  To  prove  the  Rule  for  a  Table  of  Natural  Tangents 220 

148.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Sines 221 

14!).   To  prove  the  Rule  for  a  Table  of  Logarithmic  Cosines 222 

150.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Tangents 222 

151.  Cases  of  Inapplicability  of  Rule  of  Proportional  Parts 223 

152.  Three  Methods  to  replace  the  Rule  of  Proportional  Parts. . .  224 
Examples 226 


CHAPTER   IX. 
DE  MOIVRE'S  THEOREM.  — APPLICATIONS. 

153.  De  Moivre's  Theorem 229 

154.  To  find  all  the  Values  of  (cos  0  +  V^l  sin  0)« 231 

155.  To  develop  cos  n6  and  sin  nd  in  Powers  of  sin  6  and  cos  6 233 

156.  To  develop  sin  0  and  cos  0  in  Series  of  Powers  of  0 234 

157.  Convergence  of  the  Series 235 

158.  Expansion  of  cos"  0  in  Terms  of  Cosines  of  Multiples  of  0. . .  235 

159.  Expansion  of  sin" 0  in  Terms  of  Cosines  of  Multiples  of  0. . .  236 

160.  Expansion  of  sinn  0  in  Terms  of  Sines  of  Multiples  of  0 237 

161.  Exponential  Values  of  Sine  and  Cosine 238 

162.  Gregory's  Series 239 

163.  Euler's  Series 240 

164.  Machin's  Series 241 

165.  Given  sin0  =  x  sin  (0  +  a)  ;  expand  0  in  Powers  of  x 242 

166.  Given  tan  x  =  n  tan  8  ;  expand  x  in  Powers  of  n 242 

167.  Resolve  xn  -  1  into  Factors 243 

168.  Resolve  xn  +  1  into  Factors 244 

169.  Resolve  x2n  -  2  xn  cos  0  +  1  into  Factors 245 

170.  De  Moivre's  Property  of  the  Circle 247 

171.  Cote's  Properties  of  the  Circle 248 

172.  Resolve  sin  0  into  Factors 248 

173.  Resolve  cos  e  into  Factors 250 

174.  Sum  the  Series  sin  a  +  sin(  «  +  j8)  +  etc 251 

175.  Sum  the  Series  cos  a  +  cos(a  +  j8)  +  etc 252 

176.  Sum  the  Series  sinTO  a  +  sinm(a  +  ;8)  +  etc 252 

177.  Sum  the  Series  sin  a  —  sin(«  -f  #)  -f  etc 254 

178.  Sum  the  Series  cosec  6  +  cosec  20  +  cosec  40  +  etc 254 


CONTENTS.  XI 


ART.  PAGE 

179.  Sum  the  Series  tan  6  +  |  tan  f-  +£  tan-  +  etc  ...............  255 

180.  Sum  the  Series  sin  a  +  x  sin  (  a+  #)  +  etc  .................   255 

181.  Summation  of  Infinite  Series  ............................  256 

Examples  ..............................................  257 


PART  IL 
SPHERICAL   TRIGONOMETRY. 


CHAPTER   X. 
FORMULA  RELATIVE  TO  SPHERICAL  TRIANGLES. 

182.  Spherical  Trigonometry 267 

183.  Geometric  Principles 267 

184.  Fundamental  Definitions  and  Properties 268 

185.  Formulae  for  Right  Spherical  Triangles 270 

186.  Napier's  Rules „ 272 

187.  The  Species  of  the  Parts 273 

188.  Ambiguous  Solution 274 

189.  Quadrantal  Triangles 274 

190.  Law  of  Sines 276 

191.  Law  of  Cosines 277 

192.  Relation  between  a  Side  and  the  Three  Angles 278 

193.  To  find  the  Value  of  cot  a  sin  &,  etc 279 

194.  Useful  Formulae 280 

195.  Formulae  for  the  Half  Angles 281 

196.  Formulas  for  the  Half  Sides 284 

197.  Napier's  Analogies 286 

198.  Delambre's  (or  Gauss's)  Analogies   287 

Examples 288 

CHAPTER   XI. 
SOLUTION  or  SPHERICAL  TRIANGLES. 

199.  Preliminary  Observations 297 

200.  Solution  of  Right  Spherical  Triangles 297 


xii  CONTENTS. 

ART.  PAGE 

201.  Case  I.  —  Given  the  Hypotenuse  and  an  Angle 298 

202.  Case  II.  —  Given  the  Hypotenuse  and  a  Side 209 

203.  Case  III.  —  Given  a  Side  and  the  Adjacent  Angle 300 

204.  Case  IV.  —  Given  a  Side  and  the  Opposite  Angle 301 

205.  Case  V.  —  Given  the  Two  Sides 302 

206.  Case  VI.  —  Given  the  Two  Angles 302 

207.  Quadrantal  and  Isosceles  Triangles 303 

208.  Solution  of  Oblique  Spherical  Triangles 304 

209.  Case  I.  —  Given  Two  Sides  and  the  Included  Angle 305 

210.  Case  II.  —  Given  Two  Angles  and  the  Included  Side 307 

211.  Case  III.  —  Given  Two  Sides  and  One  Opposite  Angle 309 

212.  Case  IV.  —  Given  Two  Angles  and  One  Opposite  Side 312 

213.  Case  V.  —  Given  the  Three  Sides 313 

214.  Case  VI.  —  Given  the  Three  Angles 314 

Examples 316 


CHAPTER   XII. 

THE    iN-ClRCLES    AND   Ex-ClRCLES.  —  AREAS. 

215.  The  Inscribed  Circle 324 

216.  The  Escribed  Circles 325 

217.  The  Circumscribed  Circle 326 

218.  Circumcircles  of  Colunar  Triangles 328 

219.  Areas  of  Triangles.  —  Given  the  Three  Angles 329 

220.  Areas  of  Triangles.  —  Given  the  Three  Sides 330 

221.  Areas  of  Triangles.— Given  Two  Sides  and  the  Included  Angle,  331 
Examples 332 

CHAPTER  XIII. 
APPLICATIONS  OF  SPHERICAL  TRIGONOMETRY. 

222.  Astronomical  Definitions 338 

223.  Spherical  Coordinates 339 

224.  Graphic  Representation  of  the  Spherical  Coordinates 341 

225.  Problems 342 

226.  The  Chordal  Triangle • 346 

227.  Legendre's  Theorem '. 348 

228.  Roy's  Rule 350 

229.  Reduction  of  an  Angle  to  the  Horizon 352 


CONTENTS.  xiii 

ART.  PAGE 

230.  Small  Variations  in  Parts  of  a  Spherical  Triangle 353 

231.  Inclination  of  Adjacent  Faces  of  Polyedrons 356 

232.  Volume  of  Parallelepiped 357 

233.  Diagonal  of  a  Parallelepiped 358 

234.  Table  of  Formulae  in  Spherical  Trigonometry 359 

Examples 362 


TREATISE  ON  TRIGONOMETRY 


PART  I. 
PLANE    TRIGONOMETRY. 

CHAPTER  I. 

MEASUKEMENT  OF  ANGLES, 

1.  Trigonometry  is  that  branch  of  mathematics  which 
treats  (1)  of  the  solution  of  plane  and  spherical  triangles, 
and  (2)  of  the  general  relations  of  angles  and  certain  func- 
tions of  them  called  the  trigonometric  functions. 

Plane  Trigonometry  comprises  the  solution  of  plane  trian- 
gles and  investigations  of  plane  angles  and  their  functions. 

Trigonometry  was  originally  the  science  which  treated  only  of  the 
sides  and  angles  of  plane  and  spherical  triangles ;  but  it  has  been 
recently  extended  so  as  to  include  the  analytic  treatment  of  all  theo- 
rems involving  the  consideration  of  angular  magnitudes. 

2.  The  Measure  of  a  Quantity.  —  All  measurements  of 
lines,  angles,  etc.,  are  made  in  terms  of  some  fixed  standard 
or  unit,  and  the  measure  of  a  quantity  is  the  number  of  times 
the  quantity  contains  the  unit. 

Tt  is  evident  that  the  same  quantity  will  be  represented 
by  different  numbers  when  different  units  are  adopted.  For 
example,  the  distance  of  a  mile  will  be  represented  by  the 
number  1  when  a  mile  is  the  unit  of  length,  by  the  number 
1760  when  a  yard  is  the  unit  of  length,  by  the  number  5280 
when  a  foot  is  the  unit  of  length,  and  so  on.  In  like  man- 

1 


2  PLANE  TRIGONOMETRY. 

ner,  the  r.un-.ber  expressing  the  magnitude  of  an  angle  will 
depend  on  the  unit  of  angle.      x 

EXAMPLES. 

1.  What  is  the  measure  of  2^  miles  when  a  yard  is  the 
unit  ? 

2i  miles  =  f  x  1760  yards 

=  4400  yards  =  4400  x  1  yard. 
.*.  the  measure  is  4400  when  a  yard  is  the  unit. 

2.  What  is  the  measure  of  a  mile  when  a  chain  of  66  feet 
is  the  unit  ?  Ans.  80. 

3.  What  is  the  measure  of  2  acres  when  a  square  whose 
side  is  22  yards  is  the  unit  ?  Ans.  20. 

4.  The  measure  of  a  certain  field  is  44  and  the  unit  is 
1100  square  yards ;  express  the  area  of  the  field  in  acres. 

Ans.  10  acres. 

5.  If  7  inches  be  taken  as  the  unit  of  length,  by  what 
number  will  15  feet  2  inches  be  represented  ?  Ans.  26. 

6.  If  192  square  inches  be  represented  by  the  number  12, 
what  is  the  unit  of  linear  measurement  ?        Ans.  4  inches. 

3.  Angles.  —  An  angle  is  the  opening  between  two  straight 
lines  drawn  from  the  same  point.  The  point  is  called  the 
vertex  of  the  angle,  and  the  straight  lines  are  called  the  sides 
of  the  angle. 

An  angle  may  be  generated  by  revolving  a  line  from  coin- 
cidence with  another  line  about  a  fixed  point.  The  initial 
and  final  positions  of  the  line  are  the 
sides  of  the  angle;  the  amount  of 
revolution  measures  the  magnitude  of 
the  angle ;  and  the  angle  may  be  traced 
out  by  any  number  of  revolutions  of 
the  line.  O  ~A 

Thus,  to  form  the  angle  AOB,  OB  may  be  supposed  to 
have  revolved  from  OA  to  OB  j  and  it  is  obvious  that  OB 


THE  CIRCULAR  MEASURE.  3 

may  go  on  revolving  until  it  comes  into  the  same  position 
OB  as  many  times  as  we  please ;  the  angle  AOB,  having  the 
same  bounding  lines  OA  and  OB,  may  therefore  be  greater 
than  2,  4,  8,  or  any  mlmber  of  right  angles. 

The  line  OA  from  which  OB  moves  is  called  the  initial 
line,  and  OB  in  its  final  position,  the  terminal  line.  The 
revolving  line  OB  is  called  the  generatrix.  The  point  0  is 
called  the  origin,  vertex,  or  pole. 

4.  Positive  and  Negative  Angles.  —  We  supposed  in  Art. 
3  that  OB  revolved  in  the  direction  opposite  to  that  of  the 
hands  of  a  watch.     But  angles  may,  of  course,  be  described 
by  a  line  revolving  in  the  same  direction  as  the  hands  of  a 
watch,  and  it  is  often  necessary  to  distinguish  between  the 
two  directions  in  which  angles  may  be  measured  from  the 
same  fixed  line.     This  is  conveniently  effected  by  adopting 
the  convention  that  angles  measured  in  one  direction  shall 
be  considered  positive,  and  angles  measured  in  the  opposite 
direction,  negative.     In  all  branches  of  mathematics  angles 
described  by  the  revolution  of  a  straight  line  in  the  direc- 
tion opposite  to  that  in  which  the  hands  of  a  watch  move 
are  usually  considered  positive,  and  all  angles  described  by 
the  revolution  of  a  straight  line  in  the  same  direction  as  the 
hands  of  a  watch  move  are  considered  negative. 

Thus,  the  revolving  line  OB  starts  from  the  initial  line 
OA.      When   it    revolves   in  the     B 
direction  contrary  to  that  of  the 
hands  of  a  watch,  and  comes  into 
the  position  OB,  it  traces  out  the 
positive  angle  A  OB  (marked  -f-  a)  ; 
and  when  it  revolves  in  the  same 
direction  as  the  hands  of  a  watch, 
it  traces  the  negative  angle  AOB  (marked  —  b). 

The  revolving  line  is  always  considered  negative. 

5.  The  Measure  of  Angles.  —  An  angle  is  measured  by 
the  arc  of  a  circle  whose  centre  is  at  the  vertex  of  the 


4  PLANE   TRIGONOMETRY. 

angle  and  whose  ends  are  on  the  sides  of  the  angle  (Geom., 
Art.  236). 

Let  the  line  OP  of  fixed  length  generate  an  angle  by 
revolving   in   the   positive   direction* 
round  a  fixed  point  0  from  an  initial 
position  OA.    Since  OP  is  of  constant 
length,  the  point  P  will  trace  out  the 
circumference  ABA'B'  whose  centre 
is  0.     The  two  perpendicular  diam- 
eters A  A'  and  BB'  of  this  circle  will 
inclose  the  four  right  angles  AOB, 
BOA',  A'OB',  and  B'OA. 

The  circumference  is  divided  at  the  points  A,  B,  A',  B' 
into  four  quadrants,  of  which 

AB  is  called  ihe  first  quadrant. 
BA'  «       "       "     second  quadrant. 
A'B' «       "       «     third  quadrant. 
B'A  "       "       "    fourth  quadrant. 

In  the1  figure,  the  angle  AOP1?  between  the  initial  line 
OA  and  the  revolving  line  OP1?  is  less  than  a  right  angle, 
and  is  said  to  be  an  angle  in  the  first  quadrant.  AOP2  is 
greater  than  one  and  less  than  two  right  angles,  and  is  said  to 
be  an  angle  in  the  second  quadrant.  AOP3  is  greater  than  two 
and  less  than  three  right  angles,  and  is  said  to  be  an  angle 
in  the  third  quadrant.  AOP4  is  greater  than  three  and  less 
than  four  right  angles,  and  is  said  to  be  an  angle  in  the 
fourth  quadrant. 

When  the  revolving  line  returns  to  the  initial  position 
OA,  the  angle  AOA  is  an  angle  of  four  right  angles.  By 
supposing  OP  to  continue  revolving,  the  angle  described 
will  become  greater  than  an  angle  of  four  right  angles. 
Thus,  when  OP  coincides  with  the  lines  OB,  OA',  OB',  OA, 
in  the  second  revolution,  the  angles  described,  measured 
from  the  beginning  of  the  first  revolution,  are  angles  of  five 
right  angles,  six  right  angles,  seven  right  angles,  eight  right 


THE  SEXAGESIMAL  METHOD.  5 

angles,  respectively,  and  so  on.  By  the  continued  revolu- 
tion of  OP  the  angle  between  the  initial  line  OA  and  the 
revolving  line  OP  may  become  of  any  magnitude  whatever. 

In  the  same  way  OP  may  revolve  in  the  negative  direc- 
tion about  0  any  number  of  times,  generating  a  negative 
angle  ;  and  this  negative  angle  may  obviously  have  any 
magnitude  whatever. 

The  angle  AOP  may  be  the  geometric  representative  of 
any  of  the  Trigonometric  angles  formed  by  any  number  of 
complete  revolutions,  either  in  the  positive  direction  added 
to  the  positive  angle  AOP,  or  in  the  negative  direction  added 
to  the  negative  angle  AOP.  In  all  cases  the  angle  is  said  to 
be  in  the  quadrant  indicated  by  its  terminal  line. 

There  are  three  methods  of  measuring  angles,  called 
respectively  the  Sexagesimal,  the  Centesimal,  and  the  Cir- 
cular methods. 

6.  The  Sexagesimal  Method.  —  This  is  the  method  in 
general  use.  In  this  method  the  right  angle  is  divided  into 
90  equal  parts,  each  of  which  is  called  a  degree.  Each 
degree  is  subdivided  into  60  equal  parts,  each  of  which  is 
called  a  minute.  Each  minute  is  subdivided  into  60  equal 
parts,  each  of  which  is  called  a  second.  Then  the  magni- 
tude of  an  angle  is  expressed  by  the  number  of  degrees, 
minutes,  and  seconds  which  it  contains.  Degrees,  minutes, 
and  seconds  are  denoted  respectively  by  the  symbols  °,  ',  " : 
thus,  to  represent  18  degrees,  6  minutes,  34.58  seconds,  we 
write 

18°  6'  34".58. 

A  degree  of  arc  is  ^^  of  the  circumference  to  which  the 
arc  belongs.  The  degree  of  arc  is  subdivided  in  the  same 
manner  as  the  degree  of  angle. 

Then      1  circumference  =  360°  =  21600'  =  1296000". 
1  quadrant  or  right  angle  =  90°. 

Instruments  used  for  measuring  angles  are  subdivided 
accordingly. 


PLANE  TRIGONOMETRY. 


7.  The  Centesimal  or  Decimal  Method.  —  In  this  method 
the  right  angle  is  divided  into  100  equal  parts,  each   of 
which  is  called  a  grade.     Each  grade  is  subdivided  into  100 
equal  parts,  each  of  which  is  called  a  minute.     Each  minute 
is  subdivided  into  100  equal  parts,  each  of  which  is  called 
a  second.     The  magnitude  of  an  angle  is  then  expressed  by 
the  number  of  grades,  minutes,  and  seconds  which  it  con- 
tains.    Grades,  minutes,  and  seconds  are  denoted  respect- 
ively by  the  symbols  g,  \  u :  thus,  to  represent  34  grades, 
48  minutes,  86.47  seconds,  we  write 

34*  48V  86V\47. 

The  centesimal  or  decimal  method  was  proposed  by  the  French 
mathematicians  in  the  beginning  of  the  present  century.  But 
although  it  possesses  many  advantages  over  the  established  method, 
they  were  not  considered  sufficient  to  counterbalance  the  enormous 
labor  which  would  have  been  necessary  to  rearrange  all  the  mathe- 
matical tables,  books  of  reference,  and  records  of  observations,  which 
would  have  to  be  transferred  into  the  decimal  system  before  its 
advantages  could  be  felt.  Thus,  the  centesimal  method  has  never  been 
used  even  in  France,  and  in  all  probability  never  will  be  used  in  prac- 
tical work. 

8.  The  Circular  Measure.  —  The  unit  of 
circular  measure  is  the  angle  subtended  at 
the  centre  of  a  circle  by  an  arc  equal  in 
length  to  the  radius. 

This  unit  of  circular  measure  is  called  a 
radian. 

Let  0  be  the  centre  of  a  circle  whose  radius  is  r. 
Let  the  arc  AB  be  equal  to  the  radius 


Then,  since  angles  at  the  centre  of  a 
circle  are  in  the  same  ratio  as  their 
intercepted  arcs  (Geom.,  Art.  234),  and 
since  the  ratio  of  the  circumference  of 
a  circle  to  its  diameter  is  TT  =  3.14159265 
(Geom.,  Art.  436), 


THE  CIRCULAR  MEASURE.  1 

.-.  angle  AOB  :  4rt.  angles  :  :  arc  AB  :  circumference, 

::  ?•:  2irr  :  :  1  :  ^TT. 
,.  angle  AOB  =  ^  rt  angles  =  2  rt.  angles, 

2iTT  TT 

.-.  a  radian  =  angle  AOB  =  g  ^"^  =  57°.2957795 

=  3437'.74677  =  206264".806. 

Therefore,  £/ie  radian  is  the  same  for  all  circles,  and 
=  57°.295779o. 

Let  ABP  be  any  circle  ;  let  the  angle 
AOB  be  the  radian;  and  let  AOP  be 
any  other  angle. 

Then  arc  AB  =  radius  OA. 

.-.  angle  AOP  :  angle  AOB 

: :  arc  AP  :  arc  AB  ; 
or      angle  AOP  :  radian  :  :  arc  AP  :  radius. 

.-.  angle  AOP  =  !*°A?  x  radian, 
radius 

The  measure  of  any  quantity  is  the  number  of  times  it 
contains  the  unit  of  measure  (Art.  2). 

arc  AP 

.-.  the  circular  measure  of  angle  AOP  =  —      - — 

radius 

NOTE  1.  —  The  student  will  notice  that  a  radian  is  a  little  lees  than  an  angle  of  an 
equilateral  triangle,  i.e.,  of  60°. 

Angles  expressed  in  circular  measure  are  usually  denoted  by  Greek  letters,  a,  /3, 
y,  ...,</>,  0,  $,  .... 

The  circular  measure  is  employed  in  the  various  branches  of  Analytical  Mathe- 
matics, in  which  the  angle  under  consideration  is  almost  always  expressed  by  » 
letter. 

NOTE  2.  —The  student  cannot  too  carefully  notice  that  unless  an  angle  is  obvi- 
ously referred  to,  the  letters  a,  /3, ...,  9,  </>, ...  stand  for  mere  numbers.  Thus,  n-  stands 
for  a  number,  and  a  number  only,  viz.,  3.14159  ...,  but  in  the  expression  '  the  angle 
IT,'  that  is,  '  the  angle  3.14159  ...,'  there  must  be  some  unit  understood.  The  unit 
understood  here  is  a  radian,  and  therefore  '  the  angle  n '  stands  for  '  n  radians'  or 
3.14159  ...  radians,  that  is,  two  right  angles. 

Hence,  when  an  angle  is  referred  to,  n  is  a  very  convenient  abbreviation  for 
two  right  angles. 

So  also  '  the  angle  a  or  6 '  means  '  a  radians  or  9  radians.' 

The  units  in  the  three  systems,  when  expressed  in  terms 
of  one  common  standard,  two  right  angles,  stand  thus  : 


8  PLANE  TRIGONOMETRY. 

The  unit  in  the  Sexagesimal  Method  =  —  of  2  right  angles. 

180 

"       "     "    "    Centesimal          "       =—   «  «     "          " 

200 

"       "     "    "    Circular  "        =    -    "  "     "          " 

7T 

If  D,  G,  and  6  denote  the  number  of  degrees,  grades,  a»d 
radians  respectively  in  any  angle,  then 

JUA.^  m 

180      200      TT' 

because  each  fraction  is  the  ratio  of  the  angle  to  two  right 
angles. 

9.  Comparison  of  the  Sexagesimal  and  Centesimal  Meas- 
ures of  an  Angle.  —  Although  the  centesimal  method  was 
never  in  general  use  among  mathematicians,  and  is  now 
totally  abandoned  everywhere,  yet  it  still  possesses  some 
interest,  as  it  shows  the  application  of  the  decimal  system 
to  the  measurement  of  angles. 

From  (1)  of  Art.  8  we  have 

JD_      _G 

180      200* 

.-.  D  =  ^G,  andG=  —  D. 
10  9 

EXAMPLES. 

1.   Express  49°  15'  35"  in  centesimal  measure. 
First  express   the   angle  in   degrees  and  decimals  of  a 
degree  thus  : 

60)  35" 

60)  15' .583 
49°.25972 

10 

9)  492.5972 

64*.  733024-.. . 
.-.  49°  15'  35"-  54^  73V  30V\24  .... 


COMPARISON   OF  MEASURES.  9 

2.   Express  87g  2V  25U  in  degrees,  etc. 
First  express  the  angle  in  grades  and  decimals  of  a  grade 
thus  : 

87«2V  25u=87g.0225 


78.32025 
60 


19.215 
_  60 

12.9 
.-.  87*2V25VV=78°  19'  12".9. 

Find  the  number  of  grades,  minutes,  and  seconds  in  the 
following  angles  : 

3.'    51°    4'  30".  Ans.  56*  75V    Ox\ 

4.  45°  33'    3".  50s61v20N\37. 

5.  27°  15'  46".  30*  29V  19V\75  .  •  -. 

6.  157°    4'    9".  174*52V12V\962.... 

Find  the  number  of  degrees,  minutes,  and  seconds  in  the 
following  angles  : 

7.  19*  45V  95V\  Ans.  17°  30'  48".78. 

8.  124^    5V    8V\  111°  38'  44".592. 

9.  55«  18V  35V\  49°  39'  54".54. 

10.  Comparison  of  the  Sexagesimal  and  Circular  Meas- 
ures of  an  Angle. 

From  (1)  of  Art.  8  we  have 


180 


10  PLANE  TRIGONOMETRY. 

EXAMPLES. 

1.  Find  the  number  of  degrees  in  the  angle  whose  circu- 
lar measure  is  1. 

Here  0  =  i- 

7T  2  7T 

. *^"  ^    *          O8°  *3ft'  "IfWlO 

-     22     =  ™> 

where  ^  is  used  for  TT. 

2.  Find  the  circular  measure  of  the  angle  59°  52'  30". 
Express  the  angle  in  degrees  and  decimals  of  a  degree 

thus: 

60)52.5 
59.875* 

.-.  0  =  11=11!  w  =  (.333-.)  ir  =1.0453 -. 

3.  Express,  in  degrees,  the  angles  whose  circular  measures 

7T        7T        7T        7T         2 

2'    3'    4'    6' 

NOTE  1.  — The  student  should  especially  accustom  himself  to  express  readily  in 
circular  measure  an  angle  which  is  given  in  degrees. 

4.  Express  in  circular  measure  the  following  angles  : 

60°,  22°  30',  11°  15',  270°.  Ans.  *    |,     *     ^. 

o     o     16      2 

5.  Express  in  circular  measure  3°  12',  and  find  to  seconds 
the  angle  whose  circular  measure  is  .8. 

("Take  TT  =  ~\  Ans.  ~,    45°  49'  5"T5T. 

6.  One  angle  of  a  triangle  is  45°,  and  the  circular  measure 
of  another  is  1.5.     Find  the  third  angle  in  degrees. 

Ans.  49°  5'  27"T8T. 

NOTE  2.  —  Questions  in  which  angles  are  expressed  in  different  systems  of  meas- 
urement are  easily  solved  by  expressing  each  angle  in  right  angles. 


GENERAL  MEASURE  OF  AN  ANGLE. 


11 


7.  The  sum  of  the  measure  of  an  angle  in  degrees  and 
twice  its  measure  in  radians  is  23f ;  find  its  measure  in 
degrees  (TT—  2T2). 

Let  the  angle  contain  x  right  angles. 

Then  the  measure  of  the  angle  in  degrees  =  90  x. 

"       "          "         "     "       "       "  radians  =  ^  x. 


.-.  652^=163,  .-.  x  =  -> 
4 


.-.  the  angle  is  J  of  90°  = 

8.    The  difference  between  two  angles  is  -,  and  their  sum 

is  56° ;  find  the  angles  in  degrees.  Ans.  38°,  18°. 

11.  General  Measure  of  an  Angle.  —  In  Euclidian  geom- 
etry and  in  practical  applications  of  trigonometry,  angles 
are  generally  considered  to  be  less  than  two  right  angles  ; 
but  in  the  theoretical  parts  of  mathematics,  angles  are 
treated  as  quantities  which  may  be  of  any  magnitude  what- 
ever. 

Thus,  when  we  are  told  that  an  angle  is  in  some  particu- 
lar quadrant,  say  the  second  (Art.  5),  we  know  that  the 
position  in  which  the  revolving  line  stops  is  in  the  second 
quadrant.  But  there  is  an  unlimited  number  of  angles 
having  the  same  final  position,  OP. 

The  revolving  line  OP  may  pass  from 
OA  to  OP,  not  only  by  describing  the 
arc  ABP,  but  by  moving  through  a  whole 
revolution  plus  the  arc  ABP,  or  through 
any  number  of  revolutions  plus  the  arc 
ABP. 

For  example,  the  final  position  of  OP 
may  represent  geometrically  all  the  fol- 
lowing angles  : 


12  PLANE   TRIGONOMETRY. 

Angle  AOP  =  130°,  or  360°  + 130°,  or  720°  + 130°,  or 
-  360°  +  130°,  or  -  720°  +  130°,  etc. 

Let  A  be  an  angle  between  0  and  90°,  and  let  n  be  any 
whole  number,  positive  or  negative.  Then 

(1)  2n  x  180°  +  A  represents  algebraically  an  angle  in  the 

first  quadrant. 

(2)  2n  x  180°  —  A  represents  algebraically  an  angle  in  the 

fourth  quadrant. 

(3)  (2n  -f- 1)  180°  —  A  represents  algebraically  an  angle  in 

the  second  quadrant. 

(4)  (2n  + 1)  180°  +  A  represents  algebraically  an  angle  in 

the  third  quadrant. 

In  circular  measure  the  corresponding  expressions  are 
(1)  2n7r+0,  (2)  2wir-0,  (3)  (2n  +  l)ir-0,  (4)  (2n 


EXAMPLES. 

State  in  which  quadrant  the  revolving  line  will  be  after 
describing  the  following  angles  : 

(1)  120°,    (2)  340°,    (3)  490°,    (4)  - 100°, 
(5) -380°,    (6)j7r,    (7)10*  +  ?. 

12.   Complement  and  Supplement  of  an  Angle  or  Arc.  — 

The  complement  of  an  angle  or  arc  is  the  remainder  obtained 
by  subtracting  it  from  a  right  angle  or  90°. 

The   supplement  of   an   angle   or   arc   is   the   remaiMder 
obtained  by  subtracting  it  from  two  right  angles  or  180°. 
Thus,     the  complement  of  A  is  (90°  —  A). 

The  complement  of  190°  is  (90°  -  190°)  =  -100°. 

The  supplement  of  A  is  (180°  -  A). 

The  supplement  of  200°  is  (180°  -  200°)  =  -  20°. 

The  complement  of  J?r  is  [  ?  —  *ir  i=  —  *ir. 


The  supplement  of  JTT  is  (TT  —  JTT)  =  J 


EXAMPLES.  13 

EXAMPLES. 

1.  If  192  square  inches  be  represented  by  the  number  12; 
what  is  the  unit  of  linear  measurement  ?         Ans.  4  inches. 

2.  If  1000  square  inches  be  represented  by  the  number 
40,  what  is  the  unit  of  linear  measurement  ?    Ans.  5  inches. 

3.  If  2000  cubic  inches  be  represented  by  the  number  16, 
what  is  the  unit  of  linear  measurement  ?        Ans.  5  inches. 

4.  The  length  of  an  Atlantic  cable  is  2300  miles  and  the 
length  of  the  cable  from  England  to  France  is  21  miles. 
Express  the  length  of  the  first  in  terms  of  the  second  as 
unit.  Ans. 


5.  Find  the  measure  of  a  miles  when  b  yards  is  the  unit. 

1760a 

Ans.  -      — 
b 

6.  The  ratio  of  the  area  of  one  field  to  that  of  another  is 
20  :  1,  and  the  area  of  the  first  is  half  a  square  mile.     Find 
the  number  of  square  yards  in  the  second.  Ans.  77440. 

7.  A  certain  weight  is  3.125  tons.     What  is  its  measure 
in  terms  of  4  cwt.?  Ans.  15.625. 

Express  the  following  12  angles  in  centesimal  measure  : 

8.  42°  15'  18".  Ans.  46*  95\ 

9.  63°  19'  17".  70*.35V  70V\98  .... 

10.  103°  15'  45".  114«  73v  61w  L 

11.  19°    0'18".  21*11X66V\6. 

12.  143°    9'    0".  159*    5X55V\5. 

13.  300°  15'  58".  333*  62V  90V\  1234567890. 

14.  27°  41'  51",  30«  775. 

15.  67°.4325.  74*.925. 

16.  8°  15'  27".  9*  17X  50N\ 

17.  97°    5'  15".  107g  87V  50V\ 

18.  16°  14'  19".  18*    4V29XV.... 

19.  132°    6'.  146*  77V  77v\t. 


14  PLANE  TRIGONOMETRY. 

Express  the  following  11  angles  in  degrees,  minutes,  and 
seconds : 

20.  105g  52V  75V\  Ans.  94°  58'  29".l. 

21.  82g    9V  54V\  73°  53'    9".096. 

22.  70*15V92V\  63°    8' 35".808. 

23.  15*    0V15V\  13°  30'    4".86. 

24.  154*    7V24V\  138°  39' 54".576. 

25.  324*  13V  88V\7.  291°  43'  29".9388. 

26.  10*  42V  50V\  9°  22'  57". 

27.  20*  77V  50V\  18°  41'  51". 

28.  8*  75\  7°  52'  30". 

29.  170M5V35V\  153°  24' 29  ".34. 

30.  24*    0V25V\  21°  36'    8".l. 

Express  in  circular  measure  the  following  angles : 

1453  TT 


31.  315°,  24°  13'.  Ans.  JTT, 

32.  95°  20',  12°  5'  4". 


10800 
143  w    2719  TT 


270'    40500 

33.  22i°,  1°,  57°.295.  -,    -|-,   1  radian. 

8     180 

34.  120°,  45°,  270°.  2.09439,  |,    |TT. 

35.  360°,  3-J-  rt.  angles.  27r,  JTT. 

Express  in  degrees,  etc.,  the  angles  whose  circular  meas- 
ures are : 

1  on 

36.  |TT,  ITT,  ±.  Ans.  112°.5,  120°,  --  degrees. 

2  7T 

37.  -,  -,  -•  -  degrees,  -  -  degrees,   degrees. 

463  7T  7T  7T 

38.  -,  .7854.  47°  43'  38"^,  45°. 
6 


EXAMPLES.  15 

39.  41.,  ^  2.504.     A,is.  257°  49'  43".39,  15°,  143°.468. 

40.  .0234,  1.234,  ?.  1°20'27",  70° 42' 11",  38°  11' 60". 

o 

41.  Find  the  number  of  radians  in  an  angle  at  the  centre 
of  a  circle  of  radius  25  feet,  which  intercepts  an  arc  of 
37-J-feet.  Ans.  If 

42.  Find  the  number  of  degrees  in  an  angle  at  the  centre 
of  a  circle  of  radius  10  feet,  which  intercepts  an  arc  of 
STT  feet.  Ans.  90°. 

43.  Find  the  number  of  right  angles  in  an  angle  at  the 
centre  of  a  circle  of  radius  3T2T  inches,  which  intercepts  an 
arc  of  2  feet.  Ans.  4f . 

44.  Find  the  length  of  the  arc  subtending  an  angle  of 
4|  radians  at  the  centre  of  a  circle  whose  radius  is  25  feet. 

Ans.  112-1- ft. 

45.  Find  the  length  of  an  arc  of  §0°  on  a  circle  of  4  feet 
radius.  Ans.  5f|  ft. 

46.  The  angle  subtended  by  the  diameter  of  the  Sun  at 
the   eye   of   an   observer   is  32' :    find   approximately  the 
diameter  of  the  Sun  if  its  distance  from  the  observer  be 
90  000  000  miles.  Ans.  838  000  miles. 

47.  A  railway  train  is  travelling  on  a  curve  of  half  a  mile 
radius  at  the  rate  of  20  miles  an  hour :  through  what  angle 
has  it  turned  in  10  seconds  ?  Ans.  6^  degrees. 

48.  If  the  radius  of  a  circle  be  4000  miles,  find  the  length 
of  an  arc  which  subtends  an  angle  of  1"  at  the  centre  of  the 
circle.  Ans.  About  34  yards. 

49.  On  a  circle  of  80  feet  radius  it  was  found  that  an 
angle   of  22°  30'  at  the  centre  was    subtended  by  an   arc 
31  ft.  5  in.  in  length :  hence  calculate  to  four  decimal  places 
the  numerical  value  of  the  ratio  of  the  circumference  of  a 
circle  to  its  diameter.  Ans.  3.1416. 

50.  Find  the  number  of  radians  in  10"  correct  to  four  sig- 
nificant figures  (use  ffj  for  TT).  Ans.  .00004848. 


16  PLANE  TRIGONOMETRY. 


CHAPTER   II. 
THE  TKIGONOMETKIO   FUNCTIONS, 

13.  Definitions   of    the   Trigonometric    Functions.  —  Let 
RAD  be  an  angle ;  in  AD,  one  of  the  lines  containing  the 
angle,  take  any  point  B,  and  from  B 
draw  BC  perpendicular  to  the  other  B 

line  AR,  thus  forming  a  right  triangle 
ABC,  right-angled  at  C.  Then  denot- 
ing the  angles  by  the  capital  letters  A, 
B,  C,  respectively,  and  the  three  sides  A  6  c 

opposite  these  angles  by  the  corresponding  small  italics,  a, 
b}  c*  we  have  the  following  definitions : 

a  =  opposite  side  ig  called  ^  ^  Qf  &f>  ang]e  A 
c       hypotenuse 

-  =  adjacent  side  is  called  the  cosine  of  the  angle  A. 
c       hypotenuse 

a  =  opposite  side  ig  called  the  ^          of  ^  angle  A 
b      adjacent  side 

6  =  adjacent  side  ig  called  ^  cot        nt  of  the  angle  A 
a      opposite  side 

c       hypotenuse    •        -,-,    -,  ,  -,  f  ,  i 

-  =  — J  r          —  is  called  the  secant  01  the  angle  A. 
6      adjacent  side 

c       hypotenuse  -.-.    ^  .,  T     A 

-  =  — !L-*-  —  is  called  the  cosecant  ot  the  angle  A. 
a     opposite  side 

If  the  cosine  of  A  be  subtracted  from  unity,  the  remain- 
der is  called  the  versed  sine  of  A.     If  the  sine  of  A  be  sub- 

-  The  letters  a,  b,  c  are  numbers,  being  the  number  of  times  the  lengths  of  the  sides 
contain  some  chosen  unit  of  length. 


TEIGONOMETRIC  FUNCTIONS.  17 

tracted  from  unity,  the  remainder  is  called  the  coversed  sine 
of  A ;  the  latter  term  is  hardly  ever  used  in  practice. 

The  words  sine,  cosine,  etc.,  are  abbreviated,  and  the  func- 
tions of  an  angle  A  are  written  thus  :  sin  A,  cos  A,  tan  A, 
cot  A,  sec  A,  cosec  A,  vers  A,  covers  A. 

The  following  is  the  verbal  enunciation  of  these  defini- 
tions : 

The  sine  of  an  angle  is  the  ratio  of  the  opposite  side  to  the 

hypotenuse;  or  sin  A  =  — 
c 

The  cosine  of  an  angle  is  the  ratio  of  the  adjacent  side  to 

the  hypotenuse;  or  cosA  =  — 

c 

The  tangent  of  an  angle  is  the  ratio  of  the  opposite  side  to 

the  adjacent  side;  or  tan  A  =  -• 

b 

The  cotangent  of  an  angle  is  the  ratio  of  the  adjacent  side  to 

the  opposite  side;  or  cot  A  =  — 

a 

The  secant  of  an  angle  is  the  ratio  of  the  hypotenuse  to  the 
adjacent  side;  or  sec  A  =  -• 

The  cosecant  of  an  angle  is  the  ratio  of  the  hypotenuse  to  the 

opposite  side;  or  cosec  A  =  -• 

a 

The  versed  sine  of  an  angle  is  unity  minus  the  cosine  of  the 

angle;  or  vers  A  =  1  —  cos  A  =  1  -  — 

c 

The  coversed  sine  of  an  angle  is  unity  minus  the  sine  of  the 

angle;  or  covers  A  =  1  —  sin  A  =  1 

c 

These  ratios  are  called  Trigonometric  Functions.  The 
student  should  carefully  commit  them  to  memory,  as  upon 
them  is  founded  the  whole  theory  of  Trigonometry. 

These  functions  are,  it  will  be  observed,  not  lengths,  but 


18  PLANE  TRIGONOMETRY. 

ratios  of  one  length  to  another ;  that  is,  they  are  abstract 
numbers,  simply  numerical  quantities;  and  they  remain 
unchanged  so  long  as  the  angle  remains  unchanged,  as  will 
be  proved  in  Art.  14. 

It  is  clear  from  the  above  definitions  that 

cosec  A  =  - — -,  or  sin  A  = , 

sin  A  cosec A 

sec  A     = ,  or  cosA  = 


•       V^X        \j\JiJ  J.JL    —   • 

cos  A  sec  A 

tan  A    = -,  or  cotA  = 


j         Vyj.         V/WL/J.JL. 

cot  A  tan  A 

The  powers  of  the  Trigonometric  functions  are  expressed 
as  follows  : 

(sin  A)2  is  written  sin2  A, 

(cos  A)3  is  written  cos3 A, 
and  so  on. 

NOTE.  —  The  student  must  notice  that  '  sin  A  '  is  a  single  symbol,  the  name  of  a 
number,  or  fraction  belonging  to  the  angle  A.  Also  sin2  A  is  an  abbreviation  for 
(sin  A)2,  i.e.,  for  (sin  A) x (sin  A).  Such  abbreviations  are  used  for  convenience. 

14.  The  Trigonometric  Functions  are  always  the  Same 
for  the  Same  Angle.  —  Let  BAD  be 
any  angle;  in  AD  take  P,  P',  any 
two  points,  and  draw  PC,  P'C'  per-  R, 

pendicular  to  AB.     Take  P",   any 

point  in  AB,  and  draw  P"C"  per-    ^ ,    , x 

pendicular  to  AD.  A  c   c/ 

Then  the  three  triangles  PAC,  P'AC',  P"AC"  are  equi- 
angular, since  they  are  right-angled,  and  have  a  common 
angle  at  A :  therefore  they  are  similar. 

PC  =  P'C'  =  P"C" 
'  AP~  AP'~  AP"* 

But  each  of  these  ratios  is  the  sine  of  the  angle  A.  Thus, 
sin  A  is  the  same  whatever  be  the  position  of  the  point  P  on 
either  of  the  lines  containing  the  angle  A. 


FUNCTIONS  OF  COMPLEMENTAL  ANGLES.        19 

Therefore   sin  A  is  always  the  same.     A  similar  proof 
may  be  given  for  each  of  the  other  functions. 
In  the  right  triangle  of  Art.  13,  show  that 

a  =  c  sin  A  =  c  cos  B  —  b  tan  A  =  b  cot  B, 
b  =  a  cot  A  =  a  tan  B  =  c  cos  A  =  c  sin  B, 
c  =  a  cosec  A  =  a  secB  =  b  sec  A  =  b  cosecB. 

NOTE.  —  These  results  should  be  carefully  noticed,  as  they  are  of  frequent  use  in 
the  solution  of  right  triangles  and  elsewhere. 

EXAMPLES. 

1.  Calculate  the  value  of  the  functions,  sine,  cosine,  etc., 
of  the  angle  A  in  the  right  triangles  whose  sides  a,  6,  c  are 
respectively  (1)  8,  15,  17 ;   (2)  40,  9,  41 ;  (3)  196,  315,  371 ; 
(4)  480,31,  481;   (5)  1700,  945,  1945. 

Ans.   (1)  sin  A  =  T8T,  cos  A  =  Tf,  tan  A  =  y8^,  etc. ; 

(2)  sin  A  =  £f ,  cos  A  =  ¥9T,  etc. ; 

(3)  sin  A  =  ft,  tan  A  =  |f,  etc.; 

(4)  sin  A  =  ffy,  tan  A  =  -4g8T°-,  etc. ; 

(5)  sin  A  =  fff,  tanA  =  f||-,  etc. 
In  a  right  triangle,  given  : 

2.  a  =  Vm2-f-  n2,  b  =  V2  mn ;  calculate  sin  A. 

Vra2+  n2 
Ans.  -  — 

m  -\-n 

3.  a  =  Vm2—  mn,  b  =  n;  calculate  sec  A.  m  ~  n- 

n 


vnr-\-  mn 

mn+n2 

m2-  n2 


4.  a  =  Vm2  +  mn,  c — m  -f-  n ;  calculate  tan  A.     -*/• 

5.  a  =  2  ran,  b  =  ra2—  n2 ;  calculate  cos  A. 

ra2+n2 

6.  sin  A  = -I,  c  =  200.5;  calculated.  120.3. 

7.  cos  A  =  .44,  c  =  30.5 ;  calculate  6.  13.42. 

8.  tan  A  =  JJL,  6  =  f  T ;  calculate  c.  &  Vl30. 


20 


PLANE  TRIGONOMETRY. 


15.  Functions  of  Complemental  Angles.  —  In  the  rt.  A  ABC 

we  have 

sin  A  =  -,  and  cosB  =  -.    (Art.  13.) 

c  c 

.*.  sin  A  =  cosB. 

But  B  is  the  complement  of  A,  since 
their  sum  is  a  right  angle,  or  90° ;  i.e., 


Also, 


sin  A 
cos  A 
tan  A 
cot  A 
sec  A 


=  cosB  =  cos  (90°  -A) 

=  sinB  =  sin  (90°—  A) 

=  cotB  =  cot  (90°  -A) 

=  tanB  =  tan  (90°  -A) 


a 

c 

b 

~j 
c 

a 
b' 

b 

> 

a 


=  cosecB    =cosec(90°-  A)   =  --, 


cosecA    =secB       =  sec  (90°—  A)       =  -, 

a 

vers  A     =  covers  B  =  covers  (90°  —  A)  =  1  —  -, 

c 

covers  A  =  vers  B     =  vers  (90°  -  A)     =  1  —  -. 

c 

Therefore  the  sine,  tangent,  secant,  and  versed  sine  of  an 
angle  are  equal  respectively  to  the  cosine,  cotangent,  cosecant, 
and  coversed  sine  of  the  complement  of  the  angle. 

16.  Representation  of  the  Trigonometric  Functions  by 
Straight  Lines.  —  The  Trigonometric  functions  were  for- 
merly defined  as  being  certain  straight  lines  geometrically 
connected  with  the  arc  subtending  the  angle  at  the  centre 
of  a  circle  of  given  radius. 

Thus,  let  AP  be  the  arc  of  a  circle  subtending  the  angle 
AOP  at  the  centre. 


REPRESENTATION   OF  FUNCTIONS  BY  LINES.      21 


Draw  the  tangents  AT,  BT'  meeting  DP  produced  to  T', 
and  draw  PC,  PD  _L  to  OA,  OB. 


Then       PC  was  called  the  sine  of-  the  arc  AP. 


OC 
AT 
BT' 
OT 
OT' 
AC 
BD 


cosine 

tangent 

cotangent 

secant 

cosecant 

versed  sine 

coversed  sine 


Since  any  arc  is  the  measure  of  the  angle  at  the  centre 
which  the  arc  subtends  (Art.  5),  the  above  functions  of  the 
arc  AP  are  also  functions  of  the  angle  AOP. 

It  should  be  noticed  that  the  old  functions  of  the  arc  above 
given,  when  divided  by  the  radius  of  the  circle,  become  the 
modern  functions  of  the  angle  which  the  arc  subtends  at  the 
centre.  If,  therefore,  the  radius  be  taken  as  unity,  the  old 
functions  of  the  arc  AP  become  the  modern  functions  of  the 
angle  AOP. 

Thus,  representing  the  arc  AP,  or  the  angle  AOP  by  0,  we 
have,  when  0  A  =  OP  =  1, 


22  PLANE  TRIGONOMETRY. 

smO  =^  =  ~  =  ?C, 
OP       1 


, 

OA        1 

and  similarly  for  the  other  functions. 

Therefore,  in  a  circle  whose  radius  is  unity,  the  Trigono- 
metric functions  of  an  arc,  or  of  the  angle  at  the  centre  meas- 
ured by  that  arc,  may  be  denned  as  follows  : 

The  sine  is  the  perpendicular  let  fall  from  one  extremity  of 
the  arc  upon  the  diameter  passing  through  the  other  extremity. 

The  cosine  is  the  distance  from  the  centre  of  the  circle  to  the 
foot  of  the  sine. 

The  tangent  is  the  line  which  touches  one  extremity  of  the 
arc  and  is  terminated  by  the  diameter  produced  passing  through 
the  other  extremity. 

The  secant  is  the  portion  of  the  diameter  produced  through 
one  extremity  of  the  arc  which  is  intercepted  between  the  centre 
and  the  tangent  at  the  other  extremity. 

The  versed  sine  is  the  part  of  the  diameter  intercepted 
between  the  beginning  of  the  arc  and  the  foot  of  the  sine. 

Since  the  lines  PD  or  OC,  BT',  OT  ',  and  BD  are  respect- 
ively the  sine,  tangent,  secant,  and  versed  sine  of  the  arc 
BP,  which  (Art.  12)  is  the  complement  of  AP,  we  see  that 
the  cosine,  the  cotangent,  the  cosecant,  and  the  coversed  sine  of 
an  arc  are  respectively  the  sine,  the  tangent,  the  secant,  and  the 
versed  sine  of  its  complement. 

EXAMPLES. 

1.  Prove  tan  A  sin  A  +  cos  A  =  sec  A. 

2.  "  cot  A  cos  A  +  sin  A  =  cosec  A. 

3.  "  (tan  A  -  sin  A)2+  (1  -  cos  A)2  =  (sec  A  -  I)2. 

4.  "  tan  A  -f  cot  A  =  sec  A  cosec  A. 

5.  "  (sin  A-|-  cos  A)  -5-  (sec  A  -h  cosec  A)  =  sin  A  cos  A. 

6.  "  (1  +  tanA)2+  (1  +  cot  A)2=  (sec  A  +  cosec  A)2. 


O    M 


POSITIVE  AND  NEGATIVE  LINES.  23 

7.  Given  tan  A  =  cot  2  A ;  find  A. 

8.  "      sin  A  =Cos3A;  find  A. 

9.  "      sin  A  =  cos  (45°  -  £  A)  ;  find  A. 

10.  "      tanA  =  cot6A;  find  A. 

11.  «      cot  A  =  tan  (45°+  A)  ;  find  A. 

17.  Positive  and  Negative  Lines.  — Let  AA'  and  BB'  be 

two  perpendicular  right  lines  intersecting  at  the  point  O. 
Then  the  position  of  any  point  in 
the  line  AA'  or  BB'  will  be  deter- 
mined if  we  know  the  distance  of 
the  point  from  O,  and  if  we  know 

also  upon  which  side  of  0   the    A" r- 

point  lies.  It  is  therefore  con- 
venient to  employ  the  algebraic 
signs  -+-  and  — ,  so  that  if  dis- 
tances measured  along  the  fixed 

line  OA  or  OB  from  0  in  one  direction  *  be  considered 
positive,  distances  measured  along  OA'  or  OB'  in  the  oppo- 
site direction  from  0  will  be  considered  negative. 

This  convention,  as  it  is  called,  is  extended  to  lines  parallel 
to  A  A'  and  BB';  and  it  is  customary  to  consider  distances 
measured  from  BB' towards  the  right  and  from  AA'  upwards 
as  positive,  and  consequently  distances  measured  from  BB' 
towards  the  left  and  from  AA'  downwards  as  negative. 

18.  Trigonometric  Functions  of  Angles  of  Any  Magni- 
tude. —  In   the  definitions  of  the  trigonometric  functions 
given  in  Art.  13  we  considered  only  acute  angles,  i.e.,  angles 
in  the  first  quadrant  (Art.  5),  since  the  angle  was  assumed 
to  be  one  of  the  acute  angles  of  a  right  triangle.     We  shall 
now  show  that  these  definitions  apply  to  angles  of  any  mag- 
nitude, and  that  the  functions  vary  in  sign  according  to  the 
quadrant  in  which  the  angle  happens  to  be. 


PLANE  TRIGONOMETRY. 


Let  AOP  be  an  angle  of  any  mag- 
nitude formed  by  OP  revolving  from 
an  initial  position  OA.  Draw  PM  J_ 
to  AA'.  Consider  OP  as  always 
positive.  Let  the  angle  AOP  be 
denoted  by  A  ;  then  whatever  be  the 
magnitude  of  the  angle  A,  the  defini- 
tions of  the  trigonometric  functions 
are 

MP    _.__ 

OP7 


sin  A 


OP 


cosec  A  = 


I.  When  A  lies   in   the  1st    quadrant, 
MP  is  positive  because  measured  from  M 
upwards,  OM  is  positive  because  measured 
from  0  towards  the  right  (Art.  17),  and  OP 
is  positive. 

Hence  in  the  Jirst  quadrant  all  the  func- 
tions are  positive. 

II.  When  A  lies  in  the  2d  quadrant,  as 
the   obtuse     angle   AOP,    MP    is    positive 
because  measured  from  M  upwards,  OM  is 
negative  because  measured  from  O  towards 
the  left  (Art.  17),  and  OP  is  positive. 

Hence  in  the  second  quadrant 


MP 


sin  A  =  -^—  is  positive  ; 

cos  A  =  — —  is  negative; 

MP 

tan  A  = is  negative; 

and  therefore  sec  A  and  cot  A  are  negative,  and  cosec  A  is 
positive  (Art.  13). 


FUNCTIONS   OF  ANGLES. 


III.  When  A  lies  in  the  3d  quadrant, 
as  the  reflex  angle  AGP,  MP  is  negative 
because  measured  from  M  downwards,  OM 
is  negative,  and  OP  is  positive. 

Hence  in  the  third  quadrant  the  sine, 
cosine,  secant,  and  cosecant,  are  negative, 
but  the  tangent  and  cotangent  are  positive. 

IV.  When  A  lies  in  the  4th  quadrant, 
as  the  reflex  angle  AOP,  MP  is  negative) 
OM  is  positive,  and  OP  is  positive. 

Hence  in  the  fourth  quadrant  the  sine, 
tangent,  cotangent,  and  cosecant  are  negative, 
but  the  cosine  and  secant  are  positive. 

The  signs  of  the  different  functions  are  shown  in  the 
annexed  table. 


QUADRANT. 

I. 

II. 

III. 

IV. 

Sin  and  cosec 

+ 

+ 

- 

- 

Cos  and  sec 

+ 

- 

- 

+ 

Tan  and  cot 

+ 

- 

+ 

- 

NOTE.  —  It  is  apparent  from  this  table  that  the  signs  of  all  the  functions  in  any 
quadrant  are  known  when  those  of  the  sine  and  cosine  are  known.  The  tangent  and 
cotangent  are  +  or  — ,  according  as  the  sine  and  cosine  have  like  or  different  signs. 

19.  Changes  in  the  Value  of  the  Sine  as  the  Angle  in- 
creases from  0°  to  360°.  —  Let  A  de- 
note the  angle  AOP  described  by  the 
revolution  of  OP  from  its  initial  posi- 
tion OA  through  360°.      Then,  PM 
being  drawn  perpendicular  to  AA', 
MP 


sin  A  = 


OP' 


whatever   be  the  magnitude  of  the 
angle  A. 


26  PLANE  TRIGONOMETRY. 

When  the  angle  A  is  0°,  P  coincides  with  A,  and  MP  is 
zero  ;  therefore  sin  0°  =  0. 

As  A  increases  from  0°  to  90°,  MP  increases  from  zero  to 
OB  or  OP,  and  is  positive;  therefore  sin  90°  =  1. 

Hence  in  the  1st  quadrant  sin  A  is  positive,  and  increases 
from  0  to  1. 

As  A  increases  from  90°  to  180°,  MP  decreases  from  OP 
to  zero,  and  is  positive;  therefore  sin  180°  =  0. 

Hence  in  the  2d  quadrant  sin  A  is  positive,  and  decreases 
from  1  to  0. 

As  A  increases  from  180°  to  270°,  MP  increases  from 
zero  to  OP,  and  is  negative;  therefore  sin  270°  =  —  1. 

Hence  in  the  3d  quadrant  sin  A  is  negative,  and  decreases 
algebraically  from  0  to  —  1  . 

As  A  increases  from  270°  to  360°,  MP  decreases  from  OP 
to  zero,  and  is  negative;  therefore  sin  300°=  0. 

Hence  in  the  4th  quadrant  sin  A  is  negative,  and  increases 
algebraically  from  —  1  to  0. 

20.  Changes  in  the  Cosine  as  the  Angle  increases  from  0° 
to  360°.  —  In  the  figure  of  Art.  19 


When  the  angle  A  is  0°,  P  coincides  with  A,  and 
OM  =  OP  ;  therefore  cosO°=  1. 

As  A  increases  from  0°  to  90°,  OM  decreases  from  OP  to 
zero  and  is  positive;  therefore  cos  90°=  0. 

Hence  in  the  1st  quadrant  cos  A  is  positive,  and  decreases 
from  1  to  0. 

As  A  increases  from  90°  to  180°,  OM  increases  from  zero 
to  OP,  and  is  negative;  therefore  cos  180°  =  —  1. 

Hence  in  the  2d  quadrant  cos  A  is  negative,  and  decreases 
algebraically  from  0  to  —  1. 

As  A  increases  from  180°  to  270°,  OM  decreases  from  OP 
to  zero,  and  is  negative  ;  therefore  cos  270°  =  0. 


CHANGES  IN   THE  TANGENT.  27 

Hence  in  the  3d  quadrant  cos  A  is  negative,  and  increases 
algebraically  from  —  1  to  0. 

As  A  increases  from  270°  to  360°,  OM  increases  from  zero 
to  OP,  and  is  positive;  therefore  cos  360°=  1. 

Hence  in  the  4th  quadrant  cos  A  is  positive,  and  increases 
from  0  to  1. 

21.  Changes  in  the  Tangent  as  the  Angle  increases 
from  0°  to  360°.  —  In  the  figure  of  Art.  19 


When  A  is  0°,  MP  is  zero,  and  OM  =  OP;  therefore  tan 
0°=0. 

As  A  increases  from  0°  to  90°,  MP  increases  from  zero  to 
OP,  and  OM  decreases  from  OP  to  zero,  so  that  on  both 
accounts  tan  A  increases  numerically;  therefore  tan  90°  =00. 

Hence  in  the  1st  quadrant  tan  A  is  positive,  and  increases 
from  0  to  oo. 

As  A  increases  from  90°  to  180°,  MP  decreases  from  OP 
to  zero,  and  is  positive,  OM  becomes  negative  and  decreases 
algebraically  from  zero  to  —  1  ;  therefore  tan  180°  =0. 

Hence  in  the  2d  quadrant  tan  A  is  negative,  and  increases 
algebraically  from  —  oo  to  0. 

When  A  passes  into  the  2d  quadrant,  and  is  only  just  greater  than 
90°,  tan  A  changes  from  -f  oo  to  —  oo  . 

As  A  increases  from  180°  to  270°,  MP  increases  from  zero 
to  OP,  and  is  negative,  OM  decreases  from  OP  to  zero,  and 
is  negative;  therefore  tan  270°=  oo. 

Hence  in  the  3d  quadrant  tan  A  is  positive,  and  increases 
from  0  to  oo. 

As  A  increases  from  270°  to  360°,  MP  decreases  from  OP 
to  zero,  and  is  negative,  OM  increases  from  zero  to  OP,  and 
is  positive;  therefore  tan  360°=  0. 

Hence  in  the  4th  quadrant  tan  A  is  negative,  and  increases 
algebraically  from  —  oo  to  0. 


28 


PLANE   TRIGONOMETRY. 


The  student  is  recommended  to  trace  in  a  manner  similar 
to  the  above  the  changes  in  the  other  functions,  i.e.,  the 
cotangent,  secant,  and  cosecant,  and  to  see  that  his  results 
agree  with  those  given  in  the  following  table. 

22.  Table  giving  the  Changes  of  the  Trigonometric 
Functions  in  the  Four  Quadrants. 


QUADRANT. 

I. 

II. 

III. 

IV. 

sin  varies  from 

-f 
0  to  1 

+ 
1  toO 

0  to  -  1 

-  1  to  0 

cos      "         " 

+ 
1  toO 

0  to  -  1 

-  1  toO 

+ 
.Oto  1 

tan      ".       " 

+ 
0  to  co 

—  co  to  0 

+ 
0  to  co 

-  co  toO 

cot       "         " 

+ 
oo  toO 

0  to  co 

+ 
co  to  0 

0  to  -co 

sec      "        " 

+ 
1  to  co 

—  co  to  —  1 

-  1  to  -  co 

+ 

CO    tO  1 

cosec  "        " 

+ 

CO   tO  1 

+ 
1  to  co 

-  CO   tO    -  1 

—  1   tO    —  CO 

vers     "        " 

+ 

Oto  1 

•f 

1  to  2 

+ 

2  to  1 

+ 

1  toO 

NOTE  1.  —  The  cosecant,  secant,  and  cotangent  of  an  angle  A  have  the  same  sign 
as  the  sine,  cosine,  and  tangent  of  A  respectively. 

The  sine  and  cosine  vary  from  1  to  —  1,  passing  through  the  value  0.  They  are 
never  greater  than  unity. 

The  secant  and  cosecant  vary  from  1  t«  — 1,  passing  through  the  value  oo.  They 
are  never  numerically  less  than  unity. 

The  tangent  and  cotangent  are  unlimited  in  value.  They  have  all  values  from  —  oo 
to  +00. 

The  versed  sine  and  cover sed  sine  vary  from  0  to  2,  and  are  always  positive. 

The  trigonometric  functions  change  sign  in  passing  through  the  values  0  and  oo, 
and  through  no  other  values. 

In  the  1st  quadrant  \\\e  functions  increase,  and  the  cofunctions  decrease. 

NOTE  2.  —  From  the  results  given  in  the  above  table,  it  will  be  seen  that,  if  the 
value  of  a  trigonometric  function  be  given,  we  cannot  fix  on  one  angle  to  which  it 
belongs  exclusively. 

Thus,  if  the  given  value  of  sin  A  be  \,  we  know  since  sin  A  passes  through  all 
values  from  0  to  1  as  A  increases  from  0°  to  90°,  that  one  value  of  A  lies  between  0° 


RELATIONS  BETWEEN  FUNCTIONS. 


29 


and  90°.  But  since  we  also  know  that  the  value  of  sin  A  passes  through  all  values 
between  1  and  0  as  A  increases  from  90°  to  180°,  it  is  evident  that  there  is  another 
value  of  A  between  90°  and  180°  for  which  sin  A  =  £. 

23.  Relations  between  the  Trigonometric  Functions  of 
the  Same  Angle.  —  Let  the  radius  start 
from  the  initial  position  0 A,  and  revolve 
in  either  direction,  to  the  position  OP. 
Let  0  denote  the  angle  traced  out,  and 
let  the  lengths  of  the  sides  PM,  MO, 
OP  be  denoted  by  the  letters  a,  b,  c* 

The  following  relations   are   evident 
from  the  definitions  (Art.  13)  : 


cosec  0  = 


1 


tan0  = 


sin  0 
sinfl 

COS0 


COS0' 


C0t0  = 


1 


tan0 


For 


- 
b      o     cos  0 

c 
II.    sin2  0  +  cos2  0  =  1. 

For  sin20  +  cos20  =  ^  +  ^  = 
c2      c2 


sin 


tan 


Formulae  I.,  II.,  III.,  IV.  are  very  important,  and  must  be 
remembered. 

*  a,  b,  c  are  numbers,  being  the  number  of  times  the  lengths  of  the  sides  contain 
some  chosen  uait  of  length. 


30  PLANE   TRIGONOMETRY. 

24.   Use  of  the  Preceding  Formulae . 

I.    To  express  all  the  other  functions  in  terms  of  the  sine, 

Since       sin2  0  +  cos2  0  =  1,     . :  cos  0  =  ±  Vl  —  sin20. 


COS0 

tan0 

COS0 

cosec  0  = 

sin0 

II.    To  express  all  the  other  functions  in  terms  of  the  tan- 
gent. 

Since     tan0  =  ^, 

COS0 


sec  0         Vl+tan20 


cos0  =  — !-=± 


sec  (9          Vl  +  tan20 
sec0=±  Vl 


sin  6  tan  0 

Similarly,  any  one  of  the  functions  of  an  angle  may  be 
expressed  in  terms  of  any  other  function  of  that  angle. 
The  sign  of  the  radical  will  in  all  cases  depend  upon  the 
quadrant  in  which  the  angle  6  lies. 

25.  Graphic  Method  of  finding  All  the  Functions  in  Terms 
of  One  of  them. 

To  express  all  the  other  functions  in 
terms  of  the  cosecant. 

Construct  a  right  triangle  ABC,  hav- 
ing the  side  BC  =  1.     Then 


RELATIONS  BETWEEN  FUNCTIONS. 


31 


A      AB      AB      AT3 
cosec  A  =  — —  =  -—  =  A±>. 
BC        1 


Now 


.-.  AC  =  ±  Vcosec2  A  —  1. 
sinA  =  — —  = 


—  T) 

AB      cosec  A 


A      AC       ,  Vcosec2  A  —  1 
cos  A  =  —  —  =  ±  -          —  -  , 
AB  cosec  A 


A^         Vcosec2  A  —  1 

and  similarly  the  other  functions  may  be  expressed  in  terms 
of  cosec  A. 

26.  To  find  the  Trigonometric  Functions  of  45°.  —  Let 

ABC  be  an  isosceles  right  triangle  in  which  B 

CA  =  CB. 

Then         CAB  =  CBA  =  45°. 
Let  AC  =  m  =  CB. 

Then 

AB2  =  AC2  +  CB2  =  m2  +  m2  =  2  m2. 


=  mV2. 


m 


mV2      V2 
m  1 


AB    mv2    V2 

tan  45°  =—  =  -  =  1.     cot  45° =1. 
AC      m 

sec45°=V2.     cosec45°=V2. 


27.  To  find  the  Trigonometric  Func- 
tions of  60°  and  30°. —Let  AB  be  an 

equilateral  triangle.  Draw  AD  perpen- 
dicular to  BC.  Then  AD  bisects  the 
angle  BAG  and  the  side  BC.  Therefore 
BAD  =  30°,  and  ABD  =  60°. 


32  PLANE  TRIGONOMETRY. 

Let  BA  =  2m.     .-.  BD  =  m. 

Then        AD  =  V<4  m2  —  m2  =  m  V3. 


=  ^  =  -t     -=1V3.    .-.  cosec60°=-^. 
AB       2m  y^ 

cos  60°  =»?5  =  1.        .-.  sec  60°  =  2. 

Jj  A.          *a 


.-.  cot60°  =     -. 
V3 


BD 


m 


Also 


vers60°=  1  -  cos 60°=  1  -  *  =  1. 

2      2 

sin30o  =  BD  =  ^_==l>      .    Cosec30o=2 

cos30^A5  =  ^  =  iV3.    .-.  sec30°  =  A. 
AB        2m  V3 

tan30°=~?  =  ^^:  =  ~     .-.  cotSO^VS. 

mV3      V3 

EXAMPLES. 


1.    Given    sin^  =  -;    find    the    other 
o 

trigonometric  functions. 

Let  BAG  be  the  angle,  and  BC  be 
perpendicular  to  AC.  Kepresent  BC 
by  3,  AB  by  5,  and  consequently  AC  A  4 

by  V26-9  =  4. 

A  n       A 
Then 


AB  =  5 
AC~4' 


REDUCTION   OF  FUNCTIONS.  33 


o  35 

2.  Given  sin0  =  -:  find  tan 0  and  cosec 0.  Ans.   -•>    -• 

5  4     6 

3.  Given  cos<9  =  i;  find  sin  0  and  cot  0.  |V2, -— • 

3  2v2 

4.  Given  sec0  =  4:  find  cot  0  and  sin0.  — — > -• 

Vl5      4 

5.  Given  tan0  =  V3;  find  sin  0  and  cos  9.  |V3,  -• 

6.  Given  sin  0  =  — ;  find  cos  0. 

13  lo 

7.  Given  cosec  0  =  5 ;  find  sec  0  and  tan  6. 


2V6     2V6 

8.  Given  sec  0  =  —  ;  find  sin  0  and  cot  0.  — ?    — -. 

9  41     41 

9.  Given  cot  0  =  — r ;  find  sin  0  and  sec  0.  — ->    -• 

V5  32 

10.  Given  sin  0  =  -  ;  find  cos  0,  tan  0,  and  cot  0. 

V7     3V7     V7 

T'  T'   "3"' 

11.  Given  gin  0  =  - ;  find  tan  0. 

c 


12.    Given  sin0  =     /     2;  find  tan 0. 


28.  Reduction  of  Trigonometric  Functions  to  the  1st 
Quadrant.  —  All  mathematical  tables  give  the  trigonometric 
functions  of  angles  between  0°  and  90°  only,  but  in  practice 
we  constantly  have  to  deal  with  angles  greater  than  90°. 
The  object  of  the  following  six  Articles  is  to  show  that  the 
trigonometric  functions  of  any  angle,  positive  or  negative, 
can  be  expressed  in  terms  of  the  trigonometric  functions  of 
an  angle  less  than  90°,  so  that,  if  a  given  angle  is  greater 
than  90°,  we  can  find  an  angle  in  the  1st  quadrant  whose 
trigonometric  function  has  the  same  absolute  value. 


34 


PL  A  NE   TRIG  ON  OMETli  Y. 


29.  Functions  of  Complemental  Angles.  —  Let  AA',  BB' 

be  two  diameters  of  a  circle  at  right  angles,  and  let  OP  and 
OP'  be  the  positions  of  the  radius  for 
any  angle  AOP  =  A,  and  its  comple- 
ment AOP'=  90°-  A  (Art.  12). 

Draw  PM  and  P'M'  at  right  angles 
to  OA. 

Angle  OP'M'=  BOP'=  AOP  =  A. 
Also          OP=  OP'. 

Hence  the  triangles  0PM  and  OP'M'  are  equal  in  all 
respects. 

.-.  P'M'=OM.          .-.  £!M.'=<M. 

OP'       OP 

.-.  sin  (90°  —  A)  =  cos  AOP  =  cos  A. 
Also,  OM'-PIL          .,<*'_  ™. 

.-.  cos  (90°  —  A)  =  sin  AOP  =  sin  A. 
Similarly,     tan  (90°-  A)  =  tan  AOP'  =  ?^  =  ^  =  cot  A. 

The  other  relations  are  obtained  by  inverting  the  above. 

30.  Functions  of  Supplemental  An- 
gles. —  Let  OP  and  OP'  be  the  positions 
of  the  radius  for  any  angle  AOP  =  A, 
and  its   supplement  AOP'=180°-A 
(Art.  12). 

Since  OP  =  OP',  and  POA  =  P'OA', 
the  triangles  POM  and  P'OM'  are 
geometrically  equal. 

.-.  sin  (180°-  A)  =  sin  AOP'     FM'     PM 


cos(180°-A)  =  cosAOP' 


OP 
-OM 
OP 


=  —  cos  A, 


REDUCTION   OF  FUNCTIONS. 


35 


tan  (180°-  A)  =  tan  AOP'=  =         =  _  tan  A. 

Similarly  the  other  relations  may  be  obtained. 


31.  To  prove  sin  (90° + A)  =  cos  A,  cos(90°+A)  =  -sinA, 
and  tan  (90°+  A)  =  -  cot  A. 

Let  OP  and  OP'  be  the  positions 
of  the  radius  for  any  angle  AOP  =  A, 
and  AOP'=90°+A. 

Since  OP  =  OP',  and  AOP  =  P'OB 
=  OP'M',  the  triangles  POM  and 
P'OM'  are  equal  in  all  respects. 


cos  (90°  +  A)  =  cos  AOP'=  -^  =  — ^  =  -  sin  A, 

tan  (90°+  A)  =  tan  AOP'=  £!M.'=  _2^L  =  _  cot  A. 

OM'      —  PM 


32.  Toprovesin(180°+A)  =  -si 
cos  (180°  +  A)  =  -  cos  A,  and  tan  (180° 
+  A)  =  tan  A. 

Let  the  angle  AOP  =  A;  then  the 
angle  AOP',  measured  in  the  positive 
direction,  =  (180°+ A).  p 

The  triangles  POM  and  P'OM'  are 
equal. 

.-.  sin  (180°+  A)  =sin  AOP'  =  ||£  =  ^|M  =  -sin A, 


-cos  A, 


cos  (180°+  A)  =  cos  AOP'  =  ^  = 


OP 


tan  (180°+  A)  =  tan  AOP'  = 


OM'       —  OM 


=  tan  A. 


36 


PLANE   TRIGONOMETRY. 


33.    To    prove    sin  ( —  A)  =  —  sin  A,    cos  (  —  A)  =  cos  A, 

tan  ( —  A)  =  —  tan  A. 

Let  OP  and  OP'  be  the  positions  of 
the  radius  for  any  equal  angles  AOP 
and  AOP'  measured  from  the  initial 
line  AO  in  opposite  directions.  Then 
if  the  angle  AOP  be  denoted  by  A.  the 
numerically  equal  angle  AOP'  will  be 
denoted  by  —  A  (Art.  4). 

The  triangles  POM  and  P'OM  are  geometrically  equal. 


P'M 
OP' 


^  =  -sinA, 


O 


.-.  sin(-A)  =  sinAOP'  = 


cos  (-  A)  =  cos  AOP'  =          =         =  cos  A, 
OP'       OP 


34.  To    prove   sin  (270°  +  A)  =  sin  (270°-  A)  =  -  cos  A, 

and  cos  (270°  +  A)  =  -  cos  (270°-  A)  B 

=  sin  A. 

Let  the  angle  AOP  =  A ;  then  the 
angles  AOQ  and  AOR,  measured  in 
the   positive    direction,    =(270° -A)  A'| 
and  (270°  +  A)  respectively. 

The  triangles  POM,  QON,  and  KOL 
are  geometrically  equal. 


RL  =  QN     =  OM. 


B 


RL  =  QN      -  OM 
OR      OQ         OP 


.-.  sin  (270°  +  A)  =  sin  (270°-  A)  =  -  cos  A. 


Also, 


PL  =  -  ON 
OR"    OQ 


PM 
OP' 


.-.  cos  (270°  +  A)  =  -  cos  (270°-  A)  =  sin  A. 


VALUES   OF  THE  FUNCTIONS.  37 

35.  Table  giving  the  Values  of  the  Functions  of  Any  Angle 
in  Terms  of  the  Functions  of  an  Angle  less  than  90°.  —  The 

foregoing  results,  and  other  similar  ones,  which  may  be 
proved  in  the  same  manner,  are  here  collected  for  reference. 

QUADRANT  II. 

sin  (180°-  A)  =      sin  A.  sin  (90°+  A)=      cos  A. 

cos  (180°-  A)  =  -  cos  A.  cos  (90°+  A)  =  -  sin  A. 
tan  (180°-  A)  =  -  tan  A.  tan  (90°+  A)  =  -  cot  A. 
cot  (180°-  A)  =  —  cot  A.  cot  (90°+  A)  =  —  tan  A. 
sec  (180°-  A)  =  -  sec  A.  sec  (90°+  A)  =  -  cosec  A. 

cosec  (180°  —  A)  =  cosec  A.  cosec  (90°  +  A)  =      sec  A. 

QUADRANT  III. 

sin  (180°+  A)  =  -  sin  A.  sin  (270°-  A)  =  —  cos  A. 

cos  (180°+  A)  =  —  cos  A.  cos  (270°-  A)  =  -  sin  A. 

tan  (180°+ A)  =   tan  A.  tan  (270°- A)  =   cot  A. 

cot  (180°+  A)  =   cot  A.  cot  (27(T°-A)  =  tan  A. 

sec  (180°+  A)  =  -  sec  A.  sec  (270°-  A)  =  -  cosec  A. 

cosec  (180° -f  A)  =  -  cosec  A.  cosec  (270°-  A)  =  —  sec  A. 

QUADRANT  IV. 

sin  (360°-  A)  =  -  sin  A.  sin  (270°+  A)  =  —  cos  A. 

cos  (360°- A)  =   cos  A.  cos  (270° -f  A)  =   sin  A. 

tan  (360°-  A)  =  -  tan  A.  tan  (270°+  A)  =  -  cot  A. 

cot  (360°-  A)  =  -  cot  A.  cot  (270°+  A)  =  -  tan  A. 

sec  (360°—  A)  =  sec  A.  sec  (270°+  A)  =   cosec  A. 

cosec  (360°-  A)  =  -  cosec  A.  cosec  (270°+  A)  =  -  sec  A. 

NOTE.  — These  relations*  may  be  remembered  by  noting  the  following  rules: 

When  A  is  associated  with  an  even  multiple  of  90°,  any  function  of  the  angle  is 
numerically  equal  to  the  same  function  of  A. 

When  A  is  associated  with  an  odd  multiple  of  90°,  any  function  of  the  angle  is 
numerically  equal  to  the  corresponding  cofunction  of  the  angle  A. 

The  sign  to  be  prefixed  will  depend  upon  the  quadrant  to  which  the  angle  belongs 
(Art.  5),  regarding  A  as  an  acute  angle. 

*  Although  these  relations  have  been  proved  only  in  case  of  A,  an  acute  angle, 
they  are  true  whatever  A  may  be. 


38  PLANE  TRIGONOMETRY. 

Thus,  cos  (270°  -A)  =  —  sin  A;  the  angle  270°  — A  being  in  the  3d  quadrant,  and 
its  cosine  negative  in  consequence. 
For  an  angle  in  the 

First  quadrant  all  the  functions  are  positive. 
Second  quadrant  all  are  negative  except  the  sine  and  cosecant. 
Third  quadrant  all  are  negative  except  the  tangent  and  cotangent. 
Fourth  quadrant  all  are  negative  except  the  cosine  and  secant. 

36.  Periodicity  of    the  Trigonometric  Functions.  —  Let 

AOP  be  an  angle  of  any  magnitude,  as  in  the  figure  of 
Art.  18 ;  then  if  OP  revolve  in  the  positive  or  the  negative 
direction  through  an  angle  of  360°,  it  will  return  to  the 
position  from  which  it  started.  Hence  it  is  clear  from  the 
definitions  that  the  trigonometric  functions  remain  un- 
changed when  the  angle  is  increased  or  diminished  by  360°, 
or  any  multiple  of  360°.  Thus  the  functions  of  the  angle 
400°  are  the  same  both  in  numerical  value  and  in  algebraic 
sign  as  the  functions  of  the  angle  of  400°— 360°,  i.e.,  of  the 
angle  of  40°.  Also  the  functions  of  360°+ A  are  the  same 
in  numerical  value  and  in  sign  as  those  of  A. 

In  general,  if  n-  denote  any  integer,  either  positive  or 
negative,  the  functions  of  n  x  360°+  A  are  the  same  as  those 
o/A. 

Thus  the  functions  of  1470°=  the  functions  of  30°. 

If  6  denotes  any  angle  in  circular  measure,  the  functions 
of  (2  nir  -f  6)  are  the  same  as  those  of  9.  Thus 

sin  (2  mr  -{-6)  —  sin  9,  cos  (2  mr  +  9)  =  cos  0,  etc. 

By  this  proposition  we  can  reduce  an  angle  of  any  magni- 
tude to  an  angle  less  than  360°  without  changing  the  values 
of  the  functions.  It  is  therefore  unnecessary  to  consider 
the  functions  of  angles  greater  than  360°;  the  formulae 
already  established  are  true  for  angles  of  any  magnitude 
whatever. 

EXAMPLES. 

• 

Express  sin  700°  in  terms  of  the  functions  of  an  acute 
angle. 

sin 700°=  sin  (360°+  340°)  =  sin 340°=  sin  (180°+  160°) 
=  sin  160°=  -sin  20°. 


EXAMPLES.  39 

Express  the  following  functions  in  terms  of  the  functions 
of  acute  angles  : 

1.  sin204°,  sin  510°.  Ans.   -sin24°,  sin30°. 

2.  cos  (-800°),  cos359°.  cos80°,  cosl°. 

3.  tan  500°,  tan  300°.  -tan  40°,   -cot  30°. 

Find  the  value  of  the  sine,  cosine,  and  tangent  of  the 
following  angles  : 

4.  150°.  Ans.  i     -iV3,    --  L. 

*  V3 

5.  -240°.  £V3,     -         -V3. 


6.  330°.  -|,    iV3,    --A- 

*  V3 

7.  225°.  -—  ,    -—  ,'   1. 

V2         V2 

Find  the  values  of  the  following  functions  : 

8.  sin  810°,  sin(-  240°),  cos  210°.     Ans.  1,  $V3,  -£V3. 

9.  tan  (-120°),  cot  420°,  cot  510°.  V3,    -  5_,   1. 

V3 

10.  sin  930°,  tan  6420°.  --,    i. 

2     V3 

11.  cotl035°5  cosec570°.  -1,   -2. 

37.  Angles  corresponding  to  Given  Functions.  —  When  an 
angle  is  given,  we  can  find  its  trigonometric  functions,  as  in 
Arts.  26  and  27  ;  and  to  each  value  of  the  angle  there  is 
but  one  value  of  each  of  the  functions.  But  in  the  converse 
proposition  —  being  given  the  value  of  the  trigonometric 
functions,  to  find  the  corresponding  angles  —  we  have  seen 
(Art.  36)  that  there  are  many  angles  of  different  magnitude 
which  have  the  same  functions. 

If  two  such  angles  are  in  the  same  quadrant,  they  are 
represented  geometrically  by  the  same  position  of  OP,  so 
that  they  differ  by  some  multiple  of  four  right  angles. 


40 


PLANE  TRIGONOMETRY. 


If  we  are  given  the  value  of  the  sine  of  an  angle,  it  is 
important  to  be  able  to  find  all  the  angles  which  have  that 
value  for  their  sine. 

38.  General  Expression  for  All  Angles  which  have  a 
Given  Sine  a.  —  Let  0  be  the  centre  of  the  unit  circle. 
Draw  the  diameters  A  A',  BB',  at  right 
angles. 

From  0  draw  on  OB  a  line  ON,  so 
that  its  measure  is  a. 

Through  N   draw   PP'   parallel  to  A'( 
AA'.     Join  OP,    OP',  and  draw  PM, 
P'M',  perpendicular  to  A  A'. 

Then   since    MP  =  M'P'  =  ON  =  a, 
the  sine  of  AOP  is  equal  to  the  sine  of  AOP'. 

Hence  the  angles  AOP  and  AOP'  are  supplemental  (Art. 
30),  and  if  AOP  be  denoted  by  «,  AOP'  will  =«•-«. 

Now  it  is  clear  from  the  figure  that  the  only  positive 
angles  which  have  the  sine  equal  to  a  are  a  and  TT  —  a,  and 
the  angles  formed  by  adding  any  multiple  of  four  right 
angles  to  a  and  TT  —  a.  Hence,  if  6  be  the  general  value  of 
the  required  angle,  we  have 

0  =  2mr  +  a,  or  6  =  2nir  +  ir  —  a,      .     .     .     .     (1) 

where  n  is  zero  or  any  positive  integer. 

Also  the  only  negative  angles  which  have  the  sine  equal 
to  a  are  —  (TT  +  a),  and  —  (2-n-  —  a),  and  the  angles  formed 
by  adding  to  these  any  multiple  of  four  right  angles  taken 
negatively  ;  that  is,  we  have 

(2ir  —  a),     .      .      (2) 


where  n  is  zero  or  any  negative  integer. 

Now  the  angles  in  (1)  and  (2)  may  be  arranged  thus  : 


—  a 


all  of  which,  and  no  others,  are  included  in  the  formula 


(3) 


GENERAL  EXPRESSION  FOR  ANGLES. 


41 


where  n  is  zero,  or  any  positive  or  negative  integer.  There- 
fore (3)  is  the  general  expression  for  all  angles  which  have 
a  given  sine. 

NOTE.  —  The  same  formula  determines  all  the  angles  which  have  the  same 
cosecant  as  a. 

39.  An  Expression  for  All  Angles  with  a  Given  Cosine  a.  — 

Let  0  be  the  centre  of  the  unit  circle. 
Draw  AA',  BB',  at  right  angles. 

From  0  draw  OM,  so  that  its  meas- 
ure is  a. 

Through   M   draw   PP'   parallel  to 
BB'.     Join  OP,  OP'. 

Then  since  OM  =  a,  the  cosine  of 
AOP  is  equal  to  the  cosine  of  AOP'. 

Hence,  if  AOP  =  «,  AOP'=  -  «. 

Now  it  is  clear  that  the  only  angles  which  have  the  cosine 
equal  to  a  are  a  and  —  «,  and  the  angles  which  differ  from 
either  by  a  multiple  of  four  right  angles. 

Hence  if  0  be  the  general  value  of  all  angles  whose  cosine 
is  a,  we  have 


where  n  is  zero,  or  any  positive  or  negative  integer. 

NOTE.  —  The  same  formula  determines  all  the  angles  which  have  the  same  secant 
or  the  same  versed  sine  as  a. 

40.  An  Expression  for  All  Angles  with  a  Given  Tangent  a. 

—  Let  0  be  the  centre  of  the  unit 
circle.  Draw  AT,  touching  the  circle 
at  A,  and  take  AT  so  that  its  measure 
is  a. 

Join   OT,   cutting  the    circle   at  PA' 
and  P'. 

Then  it  is  clear  from  the  figure  that 
the  only  angles  which  have  the  tan- 


P' 


gent  equal  to  a  are  a  and  TT  -f-  a,  and  the  angles  which  differ 


42  PLANE   TRIGONOMETRY. 

from  either  by  a  multiple  of  four  right  angles.     Hence  if  0 
be  the  general  value  of  the  required  angle,  we  have 


and  2mr  +  ir+a  ......     (1) 

Also,  the  only  negative  angles  which  have  the  tangent 
equal  to  a  are  —  (TT  —  a),  and  —  (2-rr  —  a),  and  the  angles 
which  differ  from  either  by  a  multiple  of  four  right  angles 
taken  negatively  ;  that  is,  we  have 


0  =  2?i7r  —  (7r-«),  and  2nir  —  (2ir  —  a),  .     .      (2) 

where  n  is  zero  or  any  negative  integer. 

Now  the  angles  in  (1)  and  (2)  may  be  arranged  thus  : 


«,   (2w  +  !)*•  +  «,   (2n  —  I)TT  +  «,   (2n  —  2)v  +  a, 
all  of  which,  and  no  others,  are  included  in  the  formula 

6  =  nir  +  a,     ............      (3) 

where  n  is  zero,  or  any  positive  or  negative  integer.  There- 
fore (3)  is  the  general  expression  for  all  angles  which  have 
a  given  tangent. 

NOTE.  —  The  same  formula  determines  all  the  angles  which  have  the  same  cotan- 
gent as  a. 

EXAMPLES. 

1.   Find  six  angles  between   —  4  right  angles  and    -f  8 
right  angles  which  satisfy  the  equation  sin  A  =  sin  18°. 
We  have  from  (3)  of  Art.  38, 

0  =  n7r  +  (-l)n—  ,  or  A  =  n  x  180°+  (-  1)W18°. 

Put  for  n  the  values  —  2,  —  1,  0,  1,  2,  3,  successively, 
and  we  get  A  =  -  360°  +  18°,  -  180°  -  18°,  18°,  180°  -  18°, 
360°  +18°,  540°  -18°; 

that  is,  -  342°,   -  198°,  18°,  162°,  378°,  522°. 

NOTE.  —  The  student  should  draw  a  figure  in  the  above  example,  and  in  each 
example  of  this  kind  which  he  works. 


TRIGONOMETRIC  IDENTITIES.  43 

2.   Find   the   four    smallest    angles   which    satisfy    the 

equations  (1)  sinA  =  i   (2)  sinA  =  — ,   (3)  sinA  =  ~, 

1  2  V2 
(4)  sinA  =  --. 

2  Ans.   (1)  30°,  150°,  -  210°,  -  330° ; 

(2)  45°,  135°,  -  225°,  -  315° ; 

(3)  60°,  120°,  -  240°,  -  300° ; 

(4)  _  30°,  _  150°,  210°,  330°. 

41.  Trigonometric  Identities. — A  trigonometric  identity 
is  an  expression  which  states  in  the  form  of  an  equation  a 
relation  which  is  true  for  all  values  of  the  angle  involved. 
Thus,  the  relations  of  Arts.  13  and  23,  and  all  others  that 
may  be  deduced  from  them  by  the  aid  of  the  ordinary 
formulae  of  Algebra,  are  universally  true,  and  are  therefore 
called  identities;  but  such  relations  as  siri0  = 
are  not  identities. 

EXAMPLES. 

1.   Prove  that     sec  0  —  tan  0  •  sin  0  =  cos  0. 


(Art.  24) 


Here        sec  9  -  tan0  sin  9  =  — '^^  sin  0        (Art.  24) 

cos  9     cos  9 

=  l-sin20 
cos0 

=  cos20 
cos0 

=  cos  9. 

2.    Prove  that     cot  9  —  sec  9  cosec  9  (1  —  2  sin2  0)  =  tan  0. 
cot0  —  sec0cosec0(l  —  2  sin20) 

=  ^  -  — *  JL  (1-  2  sin^)  (Art.  24) 

sm0      cos^   sm0 

=  cos2l9-l  +  2sm20 
sin  9  cos  0 


44  PLANE   TRIGONOMETRY. 


_  cos20  -  (sin2  0  +  cos20)  +  2  sin20          .A      2 
sin  0  cos  0 


sin  0  cos  0      cos  0 

NOTE.  —  It  will  be  observed  that  in  solving  these  examples  we  first  express  the 
other  functions  in  terras  of  the  sine  and  cosine,  and  in  most  cases  the  beginner  will 
find  this  the  simplest  course.  It  is  generally  advisable  to  begin  with  the  most  com- 
plicated side  and  work  towards  the  other. 

Prove  the  following  identities  : 

3.  cos  0  tan  0  =  sin  0. 

4.  cos  0  =  sin  0  cot  0. 

5.  (  tan  0  +  cot  0)  sin  0  cos  0  =  1. 

6.  (tan  0  -  cot  0)  sin  0  cos  0  =  sin2  0  -  cos2  0. 

7.  sin20--cosec20  =  sm4#. 

8.  sec4  0  -  tan4  0  =  sec2  0  +  tan2  0. 

9.  (sin0  —  cos0)2=l  —  2  sin  0  cos  0. 
10.  1  -  tan4  0  =  2  sec2  0  -  sec4  0. 

11. 


1  —  cos  0 

12.  (sin0  +  cos0)2+  (sin0  -  cos0)2=  2. 

13.  sin40  —  cos4  (9=  sin20  —  cos20. 

14.  sin2  0  +  vers2  0  =  2(1-  cos  0)  . 

15.  cot20-cos20  =  cot20cos20. 

EXAMPLES. 

In  a  right  triangle  ABC  (see  figure  of  Art.  15)  given  : 
1.    a=p2-\-pq,  c,=.(f-\-pq\  calculate  cot  A. 


2.    b  =  Im  -j-  w,  c  =  Z?i  -;-  m ;  calculate  cosec  A. 


EXAMPLES.  45 


3.  sin  A  =  -,  c  =  20.5  j  calculate  a.  Ans.  12.3. 

5 

4.  Given  cot  ^  A  =  tan  A ;  find  A. 

5.  "  sin  A    =cos2A;  find  A. 

6.  "  cot  A    =  tan  6  A  ;  find  A. 

7.  "  tan  A    =  cot8A;  find  A. 

8.  "  sin2A  =  cos3A;  find  A. 

9.  "       sin  A    =-;  find  cos  A  and  tan  A.       —  .    — -. 

O  '  O     '  ,  - 

6  w      V5 

10.  "       cos  A    =  -;  find  sin  A  and  tan  A  -,    -. 

5  54 

11.  «     cosec  A  =  1;  find  cos  A  and  tan  A.  2S.JL; 

3  4     V7 

12.  "       sin  A    = — r;  find  cos  A  and  tan  A. 

V3 

13.  "      cos  A   =b;  find  tan  A  and  cosec  A. 

vr=^      i 


b  Vl  -  6a 

14.  «      sin  A    =.6;  find  cos  A  and  cot  A.  -,    -. 

5    3 

15.  «       tanA    =-;  find  sin  A. 


57  V41' 

16.  «      cot  A    =  A ;  find  sec  A  and  sin  A.  A,  15. 

15  17   17 

17.  "       sin  A    =— ;  find  cos  A.  A. 

18.  "      cos  A    =  .28;  find  sin  A.  .96. 

19.  "      tanA   =^;  find  sin  A.  -. 

3  5 

20.  "       sin  A    =-;  find  cos  A. 


46  PLANE  TRIGONOMETRY. 

21.  Given  tan A=(;  find  sin  A  and  sec  A.       Ans.   -,    -• 

3  o     o 

22.  "       tanfl  =  -;  find  sin0  and  cos0. 


23.        "       cos0  =-;  find  sin 0  and  cot 0. 
a 

Va'-l 


«          Va2-l 

24.  If  sin  0  =  a,  and  tan  0  =  b,  prove  that 

(I_a2)(l"+&2)  =  l. 

Express  the  following  functions  in  terms  of  the  functions 
of  acute  angles  less  than  45° : 

25.  sin  168°,  sin 210°.  Ans.  sin  12°,   -sin 30°. 

26.  tan!25°,  tan 310°.  -cot 35°,   -cot 40°. 

27.  sec 244°,  cosec281°.  -cosec  26°,   -sec  11°. 

28.  sec  930°,  cosec  (- 600°).  -sec  30°,  sec  30°. 

29.  cot  460°,  sec  299°.  —  tan  10°,  cosec  29°. 

30.  tan  1400°,  cot  (-1400°).  -tan  40°,  cot40°. 

Find  the  values  of  the  following  functions : 

31.  sin  120°,  sin  135°,  sin  240°.       Ans.  ^    -?-,    -^. 

*        v  2  * 

32.  cos  135°,  tan  300°,  cosec  300°.  — ^  ~V3. — ~ 

V2  V3 

33.  sec315°,  cot  330°,  tan  780°.  V2,     ~V3,    V3. 

34.  sin  480°,  sin 495°,  sin  870°.  ^,    -i='    s' 

-6       V2    « 

35.  tan  1020°,  sec  1395°,  sin  1485°.  -V3,  V2,  -- 


EXAMPLES.  47 

36.  sin  (-240°),  cot(-675°),  cosec  (- 690°). 

Ana.  ^   1,   2. 

37.  cos(-300°),  cot  (-315°),  cosec  (- 1740°). 


2'         V3 

38.  tan3660°,  cos31020°.  -3V3,  -• 

8 

Find  the  value  of  the  sine,  cosine,  and  tangent  of  the 
following  angles  : 

39.  -300°.  Ans.    ^    |    VS. 

40.  -135°.  -^    -—  ,    1. 

V2         V2 


42.    -840°. 


-        V3. 


43.  1020°.  --^?,    I,     -VS. 

44.  (2n  +  l)»-  1.  ^,    -|     -VS. 

45.  (2«-l),  +  |.  -i     -^    A- 

Prove,  drawing  a  separate  figure  in  each  case,  that 

46.  sin  340°=  sin  (-160°). 

47.  sin  (-40°)  =  sin  220°. 

48.  cos  320°=  -cos  (140°). 

49.  cos  (  -  380°)  =  -  cos  560°. 

50.  cos  195°=  -cos  (-15°). 

51.  cos  380°=  -cos  560°. 


48  PLANE   TRIGONOMETRY. 

52.  cos  (-225°)  =  -cos  (-45°). 

53.  cos  1005°  =  -  cos  1185°. 

54.  Draw  an  angle  whose  sine  is  — 

a 

55.  "       "       "          "      cosecant  is  2. 

56.  "       "       "  "      tangent  is  2. 

57.  Can  an  angle  be  drawn  whose  tangent  is  427  ? 

58.  "     "       "       "       "          "       cosine  is  -  ? 

4 

59.  "     "       "       "       "  "       secant  is  7  ? 

60.  Find  four  angles  between  zero  and  -f  8  right  angles 
which  satisfy  the  equations 


(1)  sinA  =  sin20°,   (2)  sinA  =  --_,   (3)  sinA  =  --. 

V5  « 

Ans.   (1)  20°,  160°,  380°,  520°; 
,9N    5?r      7-rr      13?r      IOTT 

'      '       ~'  ~"; 


xo\    8?r      13?r      22?r      27?r 

T'  T'  T?  T' 

61.  State  the  sigrn  of  the  sine,  cosine,  and  tangent  of  each 
of  the  following  angles  : 

(1)  275°;   (2)  -91°;   (3)  -  193°  ;   (4)  -350°; 

(5)  -1000°;  (6)  2wir  +  —  • 
4 

Ans.  (1)  -,+,-;  (2)  -,-,+;  (3)  +,-,-; 
(4)  +,  +,  +;   (5)  +,+,+;   (6)  +,  -,  -. 

Prove  the  following  identities  : 

62.  (sin20-f-cos20)2=l. 

63.  (sin20  -  cos2<9)2=  1  -  4cos2^  +  4  cos4  ft 

64.  (siii0  +  cos<9)2=l  +  2sin0cos0. 

65.  (sec^  —  tan  d)  (seed  +  tan  d)  =  l. 


EXAMPLES.  49 

66.  (cosec  0  —  cot  0)  (cosec  6  -f  cot  0)  =  1. 

67.  sin3  9  +  cos3  0  =  (sin  0  +  cos  0)  (1  —  sin  0  cos  0) . 

68.  sin6  0  +  cos6  0  =  sin4  0  -f  cos4  0  -  sin2  0  cos2  0. 

69.  sin2  0  tan2  0  +  cos2  0  cot2  0  =  tan2  0  -f-  cot2  0-1. 

70.  sin0 tan20+  cosec0 sec20=  2 tan 0 sec0—  cosec 0+  sin 0. 

71.  cos30  -  sin30  =  (cos0  -  sin0)  (1  +  sin  0  cos  0). 

72.  sin6  0  +  cos6  0  =  1-3  sin2  0  cos2  0. 

73.  tan  a  -f  tan  /?  =  tan  « tan  /3  (cot  a  -f  cot  /?) . 

74.  cot «  +  tan  (3  =  cot  rc  tan  /:?  (tan  a  -f-  cot  /?) . 

75.  1  —  sin«  =  (l  -f  sina)(sec«—  tana). 

76.  1  +  cos  a  —  (1  —  cos  «)  (cosec  «  —  cot  a) . 

77.  (1  +  sin  a -j-  cos#)2=  2(1  -f-  sina)(l  -f  cos  a). 

78.  (1  —  sin«  —  cos6c)2(l  +  sina-f-  cos«)2=  4sin2«cos2«. 

79.  2  vers  a  —  vers2  a  =  sin2 «. 

80.  vers  a  (1  4-  cos  a)  =  sin2  a. 


50 


PLANE  TRIGONOMETRY. 


CHAPTER  in. 

TEIGONOMETKIO  FUNCTIONS  OF  TWO   ANGLES. 

42.  Fundamental  Formulae.  —  We  now  proceed  to  express 
the  trigonometric  functions  of  the  sum  and  difference  of 
two  angles  in  terms  of  the  trigonometric  functions  of  the 
angles  themselves. 

The  fundamental  formulae,  first  to  be  established  are  the 
following : 

sin  (x -}- y)  =  sin  x  cosy  +  cosxsiny  .  .     .     .  (1) 

cos  (x  +  y)  =  cos  x  cos  y  —  sin  x  sin  y  .  .     .     .  (2) 

sin  (x  —  y)  =  sin  x  cos  y  —  cos  x  sin  y  .  ...  (3) 

cos  (x  —  y)  =  cos  x  cosy -\-smxsiny  .  .     .     .  (4) 

NOTE.  —  Here  x  and  y  are  angles;  BO  that  (x  +  y)  and  (x  —  y)  are  also  angles. 
Hence,  sin  (x  +  y)  is  the  sine  of  an  angle,  and  is  not  the  same  as  sin  x  +  sin  y. 
Sin  (x  +  y)  is  a  single  fraction. 
Sin  x  +  siny  is  the  sura  of  two  fractions. 

43.  To  prove  that 

sin  (x  +  y)  =  sin  x  cos  y  -f  cos  x  sin  y, 
and  cos  (x  +  y)  =  cos  x  cos  y  —  sin  x  sin  y. 

Let  the  angle  AOB  =  x,  and  the 
angle  BOG  =  y ;  then  the  angle 
AOC  =  x  +  y. 

In  OC,  the  bounding  line  of  the 
angle  (x  +  y),  take  any  point  P, 
and  draw  PD,  PE,  perpendicular 
to  0  A  and  OB,  respectively ;  draw 
EH,  EK,  perpendicular  to  PD  and  OA. 


FUNDAMENTAL  FORMULA.  51 

Then  angle  EPH  =  90°-  HEP  =  HEO  =  AOE  =  x. 
DP      EK  +  PH      EK  ,  PH 


n 


=  EK  OE*.  PH  PE 
~  OE ' OP      PE ' OP 

=  since  cosy  +  coscc  sin?/. 

4.  N   OP  _  OK  -  HE  _  OK   HE 
"  OP  ~   OP     OP   OP 

_QK  OE  HE  PE 
~  OE  '  OP   PE '  OP 

=  cos  x  cosy  —  sin  a;  siny. 

NOTE.  —  These  two  formulae  have  been  obtained  by  a  construction  in  which  x  +  y 
is  an  acute  angle ;  but  the  proof  is  perfectly  general,  and  applies  to  angles  of  any 
magnitude  whatever,  by  paying  due  regard  to  the  algebraic  signs.  For  example, 


Let  AOB  =  a?,  as  before,  and 
then  AOC,  measured  in  the  positive  direc- 
tion, is  the  angle  x  +  y. 

In  OC,  take  any  point  P,  and  draw  PD, 
PE,  perpendicular  to  OA  and  OB  produced; 
draw  EH  and  EK  perpendicular  to  PD  and 
OA. 


Then,  angle  EOK  ==  180°  -x; 


=  180°-a;; 


and 


COE=y-180°. 


OE    OP     PE    OP 

=  -  sin  (180°  -  x)  cos  (y  -  180°)  +  COB  (180°  -  x)  sin  (y  -  180°) 
=  sin  x  cos  y  +  cos  x  sin  y.  (Art.  35) 

OK,  EH 


OD 
OP 


OP 


OPOP 


OK  OE   EH  PE 
OE'OP    PE'OP 


*  The  introduced  line  OE  is  the  only  line  in  the  figure  which  is  at  once  a  side  of 
two  right  triangles  (OEK  and  OEP)  into  which  EK  and  OP  enter.  A  similar 
remark  applies  to  PE. 


52 


PLANE   TRIGONOMETRY. 


=  cos  (180°  -  a?)  cos  (y  -  180°)  +  sin  (180°  -  x)  sin  (y  - 

=  cos  x  cosy  —  sin  x  sin?/.  (Art.  35) 

The  student  should  notice  that  the  words  of  the  two  proofs  are  very  nearly  the 
same. 

44.  To  prove  that 

sin  (x  —  y)  =  sin  x  cosy  —  cos  x  sin  y, 
and  cos  (x  —  y)  =  cos  x  cos  y  +  sin  x  sin  y. 

Let  the  angle  AOB  be  denoted 
by  x,  and  COB  by  y  ;  then  the  angle 
AOC  =  x-y. 

In  OC  take  any  point  P*,  and 
draw   PD,   PE,  perpendicular    to 
OA   and   OB   respectively  ;   draw 
EH,  EK,  perpendicular  to  PD  and  O 
OA  respectively. 

Then  the  angle  EPH  =  90°  -  HEP  =  BEH  =  AOB  =  x. 


OP 


OP 


- 
OP      OP 


=  EK  OE_HP  PE 
OE  '  OP     PE  '  OP 

=  sin  x  cos  y  —  cos  x  smy. 


OP 


OP 


==  EH 

OP  T  OP 


=  OK  OE     EH  PE 

OE  '  OP     PE  '  OP 

=  cos  x  cos  y  -f  sin  x  sin  y. 

NOTE  1.  —  The  sign  in  the  expression  of  the  sine  is  the  same  as  it  is  in  the  angle 
expanded  ;  in  the  cosine  it  is  the  opposite. 


*  P  is  taken  in  the  line  bounding  the  angle  under  consideration;  i.e.,  AOC. 


SINE  AND  COSINE. 


53 


NOTE  2.  —In  this  proof  the  angle  x  -  y  is  acute  ;  but  the  proof,  like  the  one  given 
in  Art.  43,  applies  to  angles  of  any  magnitude  whatever.    For  example, 

Let  AOB,  measured  in  the  positive 
direction,  =  x,  and  BOG  =  y.  Then 
AOC=x-y 

In  OC  take  any  point  P,  and  draw 
PD,  PE,  perpendicular  to  OA  and  OB 
produced:  draw  EH,  EK,  perpendicu- 
lar to  DP  and  AO  produced. 
Then, 

angle  EPH  =  EOK  =  AOB  =  360°  -  a?, 
and       POE  =  180°  -  y. 


OP 

=  EK    OE_HP    PjB 
OE  'OP     PE  '  OP 

=  sin  (360°  -  a:)  cos  (  1  80°  -  y)  -  cos  (360°  -  a;)  sin  (  180°  -  y) 
=  (—  sin  x)  (  -  cos  y)  —  cos  a;  sin  y 
=  sin  x  cos  y  —  cos  x  sin  y. 


=  _OK.  °JE_HE    PE 
OE  'OP      PE  '  OP 

=  -  cos  (3600  _  3.)  C08  (180o  _  y)  _  sin  (360o  _  x)  sin  (180o  _  y) 
=  (-  cos  a?)  (-  cos  y)  -  (-  sin  a?)  sin  y 
=  cos  x  cos  y  +  sin  x  sin  y. 

NOTE  3.  —  The  four  fundamental  formulae  just  proved  are  very  important,  and 
must  be  committed  to  memory.  It  will  be  convenient  to  refer  to  them  as  the  '  x,  y  ' 
formulae.  From  any  one  of  them,  all  the  others  can  be  deduced  in  the  following 
manner  : 

Thus,  from  cos  (x  —  y)  to  deduce  sin  (x  +  y)  .     We  have 

cos  (x  —  y)  =  cosxeosy  +  siuxsiny  .     .     .     .     (1) 
Substitute  90°—  #  for  a?  in  (1),  and  it  becomes 
cos  J90°-  (x  +  y)}  =  cos  (90°-  x)  cos?/  +  sin  (90°-  x)  siuy. 
.-.  sin  (x  -f  y  )  =  sin  x  cos  y  +  cos  x  sin  y.  (Art.  29) 

The  student  should  make  the  substitutions  indicated 
below,  and  satisfy  himself  that  the  corresponding  results 
follow  : 


54  PLANE  TRIGONOMETRY. 

From 
sin  (x  +  y)  to  deduce  cos  (x  -f  y)  substitute  (90°  4-  x)  for  x. 


cos  (x  +  y)   " 


etc. 

EXAMPLES. 

1.    To  find  the  value  of  sin  15°. 

sin  15°=  sin  (45° -30°) 

=  sin  45°  cos  30°  -  cos  45°  sin  30° 
1     V3       11 


cos  (x  —  y) 

(90°—»  for  x. 

sin  (#  -f  y) 

"          (90°+  a)  fora;. 

sin  (x  -  y) 

«          (90°  -a;)  fora;. 

cos  (a  -  ?/) 

"          —  y  for  i/. 

etc. 

etc. 

2.  Show  that  sin  75°= 

2V2 

3.  Show  that  cos  75°= 


2V2 
4.    Show  that  cos  15°= 


2V2 

O  K 

5.  If   sin  x  =  -,    and   cos  y  =  —     find     sin  (as  +  y)     and 

«£->  ^Iandf 

6.  If   sin  x  =  -,    and    cos  ?/  =  -,     find    sin  (a;  +  ?/)    and 

cos(a;—  2/).  V3 

^Ins.  1,  and  --- 

2 


FORMULA  FOR    TRANSFORMATION.  55 

45.  Formulae  for  the  Transformation  of  Sums  into  Prod- 
ucts. —  From  the  four  fundamental  formulae  of  Arts.  43 
and  44  we  have,  by  addition  and  subtraction,  the  following: 

sin  (x  +  2/)  +  sin  (x  —  y)  =  2  since  cosy  .     .     .  (1) 

sin  (x  +  y)  —  sin  (x  —  y)  =  2  cos  x  sin  y  ...  (2) 

cos  (x  +  y)  +  cos  (x  —  ?/)  =  2  cos  x  cos  y  .     .     .  (3) 

cos  (x  —  y)  —  cos  (x  +  y)  =  2  since  siny  .     .     .  (4) 

These  formulae  are  useful  in  proving  identities  by  trans- 
forming products  into  terms  of  first  degree.  They  enable 
us,  when  read  from  right  to  left,  to  replace  the  product  of  a 
sine  or  a  cosine  into  a  sine  or  a  cosine  by  half  the  sum  or 
half  the  difference  of  two  such  ratios. 

Let  x-\-y  =  A,  and  x  —  y  =  B. 


,  and  y  =  -J(A-B). 

Substituting  these  values  in  the  above  formulae,  and 
putting,  for  the  sake  of  uniformity  of  notation,  x,  y  instead 
of  A,  B,  we  get 

sin  x  +  sin  y  =  2  sin  %(x-\-y)  cos  ^  (#  —  ?/)  .  .  (5) 

sin  x  —  sin  y  =  2  cos  ^  (x  +  y)  sin  ^-  (#  —  y)  .  .  (6) 

cos  x  -f  cos  y  =  2  cos  i  (a;  -f-  y)  cos  £  (a?  —  y)  .  .  (7) 

cos  ?/  —  cos  #  =  2  sin  £  (a;  +  y)  sin  £  (a:  —  y)  .  .  (8) 

The  formulae  are  of  great  importance  in  mathematical 
investigations  (especially  in  computations  by  logarithms)  ; 
they  enable  us  to  express  the  sum  or  the  difference  of  two 
sines  or  two  cosines  in  the  form  of  a  product.  The  student 
is  recommended  to  become  familiar  with  them,  and  to  coirx 
mit  the  following  enunciations  to  memory  : 

Of  any  two  angles,  the 

Sum  of  the  sines  =  2  sin  £  sum  -cos^diff. 
Diff.  "    "       "      =2cosjsum.sinidiff. 


56  PLANE  TRIGONOMETRY. 

Sum  of  the  cosines  =  2  cos  ^  sum  •  cos  %  diff. 
Diff.  "    "        «       =  2  sin  £  sum- sin  £  diff. 

EXAMPLES. 

1.  sin  5  a  cos  3  #  =  £  (sin  8  # -f-  sin2#). 

For,         sin5cccos3a;=  Jjsin  (5x  +  3x)  +  sin  (5x  —  3x)  \ 
=  \  (sin  8  x  +  sin  2  x) . 

2.  Prove  sin 0 sin 30         =£  (cos 2(9  -  cos 40). 

3.  "  2sin0cos<£        =  sin  (0  +  <£)  +  sin  (6  —  <£). 

4.  "  2sin20cos3<£  =  sin(20-f  3<£)  +  sin  (20— 3<£). 

5.  «  sin  60°  +  sin  30°  =  2  sin  45°  cos  15°. 

6.  «  sin 40° -sin  10°  =2 cos 25°  sin  15°. 

7.  "  sinl00+sin60=2sin80cos20. 

8.  "  sin8«—  sin4a  =  2cos6asin2a. 

9.  "  sin  3  x  +  sin  a;    =  2  sin  2  a;  cos  x. 

10.  "      sin  3  »  — sin  a;    =  2  cos  2  a;  sin  x. 

11.  "      sin  4  07  +  sin  2  a;  =  2  sin  3  x  cos  a;. 

46.  Useful  Formulae.  —  The  following  formulae,  which 
are  of  frequent  use,  may  be  deduced  by  taking  the  quotient 
of  each  pair  of  the  formulae  (5)  to  (8)  of  Art.  45  as  follows : 

^     since -f-  sin?/_  2 sin %(x  -\-y)  cos^(x—,y) 
sina;  —  smy      2cos£(#  +  y)  sin|(x  —  y) 

=  tan  $(x  -+-?/)  cot  %(x  —  y) 

).  (Art.  24) 

—  y) 

The  following  may  be  proved  by  the  student  in  a  similar 
manner : 

2.   !H£+liM 
cos  x  -f  cos  y 


THE  TANGENT  OF  TWO  ANGLES.  57 

3.   ""s  +  'foy-cotiCs-y), 

cos  y  —  cos  a; 


4.  -          =  tani  (g-y), 

COSX  +  COS?/ 


5. 


cos  y  —  cos  a; 

6. 

cosy  —  cos  a; 

47.  The  Tangent  of  the  Sum  and  Difference  of  Two  Angles. 

—  Expressions  for  the  value  of  tan(#  +  y),  tan  (x  —  y),  etc., 
may  be  established  geometrically.  It  is  simpler,  however, 
to  deduce  them  from  the  formulae  already  established,  as 
follows  : 

Dividing  the  first  of  the  '  x,  y  '  formulae  by  the  second, 
we  have,  by  Art.  23, 

tan  (x  +  y)  =  sin  (x  +  y)  =  sin  a?  cos  y  +  cos  g  sin  y 
cos(ie-fy)      cos  x  cos  y  —  sin  x  sin  y 

Dividing  both  terms  of  the  fraction  by  cos  x  cos  y, 

sin  x  cos  y      cos  a;  sin  y 

cos  a;  cos  y     cos  a?  cos?/ 

tan  (x  +  y)  =  -  2  -  .  -  __2 

cos  a;  cosy      sin  a;  sin  y 

cos  a;  cosy      cos  a;  cos  a; 

=  tana  +  tany  1} 

1  —  tana;tany  v 

In  the  same  manner  may  be  derived 


tan(s-y)=  ~  (2) 

1-ftanxtany 

Also,    cot(a?  +  y)  =  COta?COty""1  (3) 

cot  x  +  cot  y 


and          eot(x-y)  =  (4) 

coty  —cot  a; 


58  PLANE  TRIGONOMETRY. 


EXERCISES. 
Prove  the  following  : 


1  —  tana; 

2.  tan(s-45')  = 

1  +  tana; 

3  sin  (x  +  y)      _  tan  x  -f  tan  y 
sin  (a;  —  y)          tan  x  —  tan  ?/ 

4  cos  (a;  —  y)     _  tan  x  tan  y  -f  1 
cos  (z  -f  ?/)         1  —  tan  x  tan  y 

5.  sin  (x  +  y)  sin  (x  —  y)  =  sin2  a;  —  sin2?/ 

=  cos2?/  —  cos2  a;. 

6.  cos  (x  +  y)cos(x  —  y)  =  cos2x  —  sin2?/ 

=  cos2?/  —  sin2  a?. 

7.  tail  x  ±  tan  y  =>"(***). 

cos  a;  cos?/ 

8.  cotx±coty  = 

sn  a;  sn  y 

sin  2  a;     cos2a; 

9.  —  =  sec  x. 
sin  x        cos  x 

10.  If  tan  a;  =  £  and  tan  ?/  =  J,  prove  that  tan  (a;  -f  y)  =  f  , 
and  tan  (a;  —  y)  =  -J. 

11.  Prove  that  tan  15°  =  2  -  V3. 

12.  If  tan  x=  f  ,  and  tan  y  =  Jy,  prove  that  tan  (x  +  y)=  1. 
What  is  (x  H-  y)  in  this  case  ? 

48.  Formulae  for  the  Sum  of  Three  or  More  Angles.  —  Let 

x}  y,  z  be  any  three  angles  ;  we  have  by  Art.  43, 


EXAMPLES.  59 

sin  (x  +  y  -f  z)  =  sin  (a;  -|-  y)  cos 2  -f  cos  (a;  -f  y)  sinz 
=  sin  ic  cos  y  cos  2  -f-  cos  x  sin  ?/  cos  z 
+  cos  x  cosy  sin z  —  sin  a;  sin?/  sinz   .     .   (1) 

In  like  manner, 

cos  (x  +  y  +  z)  =  cos  a;  cos  y  cos  z  —  sin  x  sin  ?/  cos  z 

—  sin  a;  cosy  sinz  —  cos  x  sin  y  sinz   .     .   (2) 

Dividing  (1)  by  (2),  and  reducing  by  dividing  both  terms 
of  the  fraction  by  cos  x  cosy  cos  0,  we  get 

tanfff  +  sH-g)  =  tana?  +  tany  +  tans  -  tana?  tany  tanz      ^ 
1  —  tana;  tan  ?/  —  tany  tanz  —  tanz  tana; 

EXAMPLES. 

1.  Prove  that  sin  x  +  sin  y  +  sin  z  —  sin  (a;  -f  y  +  z) 

=  4sin  JO  -f  y)  sin$(y  +  z)  sin  J(z  +  JB). 
By  (6)  of  Art.  44  we  have 

sin  a;—  sin  (a;  -f  y  -f  z)  =  —  2cos  J(2#  +  y  -f  z)  sin  J(?/ +  z), 
and  siuy  -\- sinz  =  2 sin ^(y  +  2)  cos^(?/  —  z). 

.  •.  sin  a;  +  sin  y  -f-  sin  z  —  sin  (a;  +  y  -f-  2) 

=  2sin£(#+z)  cos^(2/— z)  — 2cos^(2a;+2/+z)  sin 
=2sin|-(2/  +  z)  |cosj(y  —  z)  —  cos %(2x  +  y  +  z)  £ 
in^(a;4-  y)  sin  J(a?  +  z) 
J(2/  +  z)  sin|(z  +  «). 
Prove  the  following : 

2.  cos  a;  +  cos  y  +  cos  z  +  cos  (a;  +  y  +  z) 

-f  z)  cos£(z  +  a?)  cos  J 


60  PLANE  TRIGONOMETRY. 

3.  sin  (x  +  y  —  z)  =  sin  x  cos  y  cos  z  +  cos  cc  sin  i/  cos  z 

—  cos  a;  cosy  simz  -f  sin  a?  siny  sinz. 

4.  sin  x  +  sin  i/  —  sin  z  —  sin  (cc  -f  y  —  z) 

=  4  sin  $(x  —  z)  sinj(;z/  —  2)  s 

5.  sin  (y  —  z)  +  sin  (2  —  a?)  -f-  sin  (a;  —  y) 

—  2;)  sin  ^(2  —  x)  sin  *  (x  —  t/)  =  0. 


49.  Functions  of  Double  Angles.  —  To  express  the  trigo- 
nometric functions  of  the  angle  2  a?  in  terms  of  those  of  the 
angle  x. 

Put  y  =  x  in  (1)  of  Art.  42,  and  it  becomes 
sin  2  x  =  sin  x  cos  a;  +  cos  x  sin  a;, 
or  sin  2  ie  =  2  sin  x  cos  a;      .......     (1) 

Put  y  =  x  in  (2)  of  Art.  42,  and  it  becomes 

cos  2  x  =  cos2  x  —  sin2  a;  .......     (2) 

=  l-2sin2a;  ........     (3) 

or  =2cos2#-l  ........     (4) 

Put  y  =  x  in  (1)  and  (3)  of  Art.  47,  and  they  become 

tan2a;=    2tana;      ........     (5) 

1  —  tan2  a 


(6) 


2  cot  a 
Transposing  1  in  (4),  and  dividing  it  into  (1),  we  have 

sin2*     =  tan*.  (7) 


EXAMPLES.  61 

NOTE.  —  These  seven  formulae  are  very  important.     The  student  must  notice  that 
x  is  any  angle,  and  therefore  these  formulas  will  be  true  whatever  we  put  for  x. 

Thus,  if  we  write  -  for  x,  we  get 

sin  a;  =  2  sin?  cos-     ...  ........     (8) 


--6in2- 
2  2 


or  =  l-28in2-  =  2cos2--l  .  .  (10) 

2  2 

and  so  on. 

EXAMPLES. 


Prove  the  following  : 
1.   2cosec2x     =  sec#cosec#. 


2.    __   =8ec2«. 
cosecx  —  2 


3. 


1  +  tan2  x 
4.    l' 


5.  tan#+cot#  =  2cosec2x. 

6.  cotx  —  tan  «  =  2  cot  2  a;. 

sina; 


+  oos  a; 

sinz 
1  —  coso; 


9.    Given  sin  45°=-L;  find  tan  22f.          Ans.   V2-1. 
V2 

10.    Given  fean#  =  -;  find  tan  2  a;,  and  sin  2  a.          —  ,    ?|- 
4  7      25 

50.  To  Express  the  Functions  of  3  a;  in  Terms  of  the  Func- 
tions of  x. 

Put  y  =  2#  in  (1)  of  Art.  42,  and  it  becomes 


62  PLANE  TRIGONOMETRY. 

=  sin  (2#-f  x) 

=  sin2a;  cosse  +  cos2x  sin  a; 

=  2  sin  a;  cos2  a;  +  (1  —  2  sin2  a)  sin  X     (Art.  49) 

=  2sinic(l  —  sin2  a?)  -f-  sinx  —  2  sin3  a; 

=  3  sin  a;  —  4sin3#. 


=  cos  2  x  cos  x  —  sin  2  x  sin  x 

=  (2cos2#—  1)  cos  a;  —  2sin2iccosx     (Art.  49) 

=  4cos3ic  — 


tan2a;4-tana; 
1  —  tan  2  x  tan  x 

2  tana; 


1  —  tan2  a; 


-f  tanx 


1_     2  tan2  a; 


1  —  tan2  a; 

3  tan  a;  —  tan3  a; 
1  —3  tan2  a; 

EXAMPLES. 


Prove  the  following  : 


2. 


sma; 

Sin3a:-sina; 
cos  3  x  +  cos  x 

3.   sin3x  +  cos* 
cos  x  —  sin  x 


4. 


FUNCTIONS  OF  THE  HALF  ANGLE.  63 

-  1  -  +  _  - 

tan  3x  —  tana;      cot#  —  cot3# 


1  —  COS  X 


51.   Functions  of  Half  an  Angle.  —  To  express  the  func- 
tions of  -  in  terms  of  the  functions  of  x. 

Since  cosx=  1  —  2  sin2-, 


or 

.      9  X 

sin 

=  2cos'|-l     . 
1  —  cos  x 

.     .    [Art.  49,  (10)] 

and 

.  .     0111 

cos2*3 

1  -f-  cos  x 

(2} 

2 

2 

Or 

sinx 

/I  —  COS  X 

(3\ 

2 

-  V       2 

and 

cosx 

/I  -f  cos  a; 

2 

-\        2 

Bv 

division    tan 

/I  —  cos  aj 

1  —  cos  a:              x-v 

*->y 

2i 

*"  \1  -f-  cos  a; 

sinaj 

By  formulae  (3),  (4),  and  (5)  the  functions  of  half  an 
angle  may  be  found  when  the  cosine  of  the  whole  angle  is 
given. 

52.  If  the  Cosine  of  an  Angle  be  given,  the  Sine  and  the 
Cosine  of  its  Half  are  each  Two-Valued. 

By  Art.  51,  each  value  of  cos  a  (nothing  else  being  known 

about  the  angle  x)  gives  two  values  each  for  sin-  and  cos-, 

2  £ 

one  positive  and  one  negative.     But  if  the  value  of  x  be 


64 


PLANE  TRIGONOMETRY. 


COS- 

2 


IS 


given,  we  know  the  quadrant  in  which  -  lies,  and  hence 
we  know  which  sign  is  to  be  taken. 

Thus,  if  x  lies  between  0°  and  360°,  -  lies  between  0°  and 
180°,  and  therefore  sin-  is  positive;  but  if  x  lies  between 
360°  and  720°,  ?  lies  between  180°  and  360°,  and  hence 

£i 

sin-  is  negative.     Also,  if  x  lie  between  0°  and  180°,  cos^ 
•"  2 

is  positive;    but  if  x  lie  between  180°  and  360 

negative. 

The  case  may  be  investigated  geometrically  thus : 

Let  OM  =  the  given  cosine  (radius  being  unity,  Art.  16), 

=  cos  x.     Through  M  draw  PQ  per- 
pendicular to  0 A ;  and  draw  OP,  OQ. 

Then   all   angles  whose  cosines  are 

equal  to  cos  a;  are  terminated  either 

by  OP  or  OQ,  and  the  halves  of  these 

angles  are  terminated  by  the  dotted 

lines  Op,  Oq,  Or,  or  Os.     The  sines 

of  angles  ending  at  Op  and  Oq  are 

the  same,  and  equal  numerically  to 

those  of  angles  ending  at  Or  and  Os ;  but  in  the  former  case 

they  are  positive,  and  in  the  latter,  negative;    hence  we 

obtain  two,  and  only  two,  values  of  sin  -  from  a  given  value 
of  cos  x. 

Also,  fche  cosines  of  angles  ending  at  Op  and  Os  are  the 
same,  and  have  the  positive  sign.  They  are  equal  numeri- 
cally to  the  cosines  of  the  angles  ending  at  Og  and  0?*,  but 
the  latter  are  negative ;  hence  we  obtain  two,  and  only  two, 

values  of  cos-  from  a  given  value  of  cosx. 

2 

Also,  the  tangent  of  half  the  angle  whose  cosine  is  given 
is  two-valued.  This  follows  immediately  from  (5)  of  Art. 
51. 


FUNCTIONS   OF  THE  HALF  ANGLE. 


65 


53.   If  the  Sine  of  an  Angle  be  given,  the  Sine  and  the 
Cosine  of  its  Half  are  each  Four- Valued. 


2sin-cos-  =  sina;    .     .     .     (Art.  49) 

w  £ 


We  have 
and 

By  addition,        [sin  f  +  cos  ?  ]  =14-  sin  a;. 


sin2- 4- cos2- =1     ....     (Art.  23) 

2  2 


/  x  x\2 

By  subtraction,  I  sin-  —  cos-  j  =  1  —  sin  a;. 


.-.  sin|  +  cos| 


.     .     (1) 


and 


and 


sin--  cos-  =±  Vl-sinz    .     .     (2) 


.-.  2sin^=  ± 

2 


sinx  ±  Vl  —  sina    .     .     (3) 


2cos-=  ±  Vl  +  since  =F  Vl  —  sin  a;     .     .     (4) 

a 


Thus,  if  we  are  given  the  value  of  sin  a;  (nothing  else 
being  known  about  the  angle  x),  it  follows  from  (3)  and 

(4)  that  sin-  and  cos-  have  each  four  values  equal,  two 

2  2 

by  two,  in  absolute  value,  but  of  contrary  signs. 
The  case  may  be  investigated  geometrically  thus : 
Let  ON  =  the  given  sine    (radius  being  unity)  =  sin  x. 

Through  N  draw  PQ  parallel  to  OA ; 

and  draw  OP,  OQ.     Then  all  angles 

whose   sines   are   equal  to  sin  a?  are 

terminated  either  by  OP  or  OQ,  and 

the  halves  of  these  angles  are  termi- 
nated by  the  dotted  lines  Op,  Og,  Or, 

or  Os.     The  sines  of  angles  ending 

at  Op,  Oq,  Or,  and  Os  are  all  different 


Q 


66  PLANE  TRIGONOMETRY. 

in  value  ;  and  so  are  their  cosines.     Hence  we  obtain  four 

T*  'T* 

values  for  sin-,  and  four  also  for  cos-,  in  terms  of  x. 

—  Jj 

When  the  angle  x  is  given,  there  is  no  ambiguity  in  the 
calculations  ;  for  -  is  then  known,  and  therefore  the  signs 

and  relative  magnitudes  of  sin-  and  cos^  are  known.    Then 

2  2 

equations  (1)  and  (2),  which  should  always  be  used,  im- 
mediately determine  the  signs  to  be  taken  in  equations  (3) 
and  (4). 

Thus,  when  -  lies  between  —  45°  and  4  45°,  cos  -  >  sin  -, 

2  22 

and  is  positive. 

Therefore  (1)  is  positive,  and  (2)  is  negative  •  and  hence 
(3)  and  (4)  become 

2  sin  -  =  Vl  4-  sin  x  —  Vl  —  sin  x, 

2 

2  cos  -  =  VT  4-  sin  x  +  Vl  —  sin  x. 


When  -  lies  between  45°  and  135°,  sin  ->  cos-,  and  is 

2  22 

positive. 

Therefore  (1)  and  (2)  are  both  positive;  and  hence  (3) 
and  (4)  become 

2  sin-  =  Vl  4-  sin  a;  4-  Vl  —  sin#, 

2 

2  cos  -  =  Vl  4-  sin  x  —  Vl  —  sin  x. 
And  so  on. 

54.  If  the  Tangent  of  an  Angle  be  given,  the  Tangent 
of  its  Half  is  Two-Valued. 

2  tan   i 
We  have  tan0  =  -  1    .....   (Art.  49) 


FUNCTIONS   OF  THE  HALF  ANGLE.  67 


Put  tan-  =  #;  thus 

2 


tan0 


. 

2  tan  6 

£ 

Thus,  given  tan0,  we  find  £«;o  unequal  values  for  tan-, 
one  positive  and  one  negative. 

This  result  may  be  proved  geometrically,  an  exercise  which 
we  leave  for  the  student. 

55.  If  the  Sine  of  an  Angle  be  given,  the  Sine  of  One- 
Third  of  the  Angle  is  Three-  Valued. 

We  have  sin  3  x  =  3  sin  x  —  4  sin3  x    .     .     (Art.  50) 

f\ 

Put  x  =  -,  and  we  get 

3 


3  3' 

a  cubic  equation,  which  therefore  has  three  roots. 

EXAMPLES. 

1.   Determine  the  limits  between  which  A  must  lie  to 
satisfy  the  equation 

2  sin  A  =  —  Vl  +  sin  2  A  —  Vl  —  sin  2  A. 

By  (1)  and  (2)  of  Art.  53,  2  sin  A  can  have  this  value 
only  when 

sin  A  +  cos  A  =  —  Vl  +  sin  2  A, 


and  sin  A  —  cosA  =  —  Vl  —  sin2A; 

i.e.,  when  sin  A  >  cos  A  and  negative. 


68  PLANE  TRIGONOMETRY. 

Therefore  A  lies  between  225°  and  315°,  or  between  the 
angles  formed  by  adding  or  subtracting  any  multiple  of 
four  right  angles  to  each  of  these ;  i.e.,  A  lies  between 

2tt7r  +  —  and  2mr  +  1— , 
4  4 

where  n  is  zero  or  any  positive  or  negative  integer. 

2.  Determine  the  limits  between  which  A  must  lie  to 
satisfy  the  equation 


2  cos  A  ==  Vl  +  sin  2  A  —  Vl  —  sin  2  A. 

By  (1)  and  (2)  of  Art.  53,  2  cos  A  can  have  this  value 
only  when 

cos  A  +  sin  A  =  Vl  -f-  sin  2  A, 


and  cos  A  —  sin  A  =  —  Vl  —  sin2A; 

i.e.,  when  sin  A  >  cos  A  and  positive. 
Therefore  A  lies  between 

2W7T  +  -  and  2tt7r  +  ^", 
4  4 

where  n  is  any  positive  or  negative  integer. 

3.  State  the  signs  of  (sin  6  +  cos  0)   and   (sin  0  —  cos  0) 
when   0   has   the   following   values:    (1)    22°;     (2)    191°; 
(3)  290°;   (4)  345°;   (5)   -22°;   (6)   -275°;   (7)    -470°; 
(8)  1000°. 

Am.   (1)  +,  -;   (2)  -,  +;   (3)  -,  -;   (4)  +  ,  -; 
(5)  +,  -;   (6)  +,  +;   (7)  -,  -;   (8)  -,  -. 

4.  Prove  that  the   formulae   which  give  the  values  of 

sin-  and  of  cos^  in  terms  of  sin  a;  are  unaltered  when  x 

2i  — ' 

has  the  values 

(1)  92°,  268°,  900°,  4wir  +  Jir,  or  (4n  +  2)  TT-  |TT; 

(2)  88°,  -  88°,  770°,  -  770°,  or  4n7r  ±  \ 


VALUES  OF  SPECIAL  ANGLES. 

5.   Find  the  limits  between  which  A  must  lie  when 


2sinA=  Vl  +  sin2A  —  Vl  —  sin2A. 

56.  Find  the  Values  of  the  Functions  of  22|°.  — In  (3), 
(4),  and  (5)  of  Art.  51,  put  x  =  45°.     Then 


cos  22i°=  A  /I  +  cos  45°      \/2  +  V2 

2        ~~ 


sin  45° 

Since  22£°  is  an  acute  angle,  its  functions  are  all  positive. 
The  above  results  are  also  the  cosine,  sine,  and  cotangent 
respectively  of  67£°,  since  the  latter  is  the  complement  of 
(Art.  15). 


57.  Find  the  Sine  and  Cosine  of  18°. 
Let  cc=18°;  then  2  a;  =  36°,  and  3  x  =  54°. 

.-.  2x  +  3z  =  90°. 

.-.  sin  2  x  =  cos  3  x    .....     (Art.  15) 
.•.  2  sin  x  cos  x  =  4  cos3  x  —  3  cos  x  .     .     (Art.  50) 
or  2sinjc  =  4cos2ic  —  3 

=  1  —  4  sin2  x. 

Solving  the  quadratic,  and  taking  the  upper  sign,  since 
sin  18°  must  be  positive,  we  get 

sin  18°=^*. 


+  2 


Also,    cos  18°=  Vl  -  sin218°= 

4 

Hence  we  have  also  the  sine  and  cosine  of  72°  (Art.  15). 


70  PLANE  TRIGONOMETRY. 

56.  Find  the  Sine  and  Cosine  of  36°. 

cos36°=l-2sin218°     .     .     .    [(3)  of  Art.  49] 


A/10  -  2  V5 


.-.  sin  36°=  Vl  -  cos2  36°  = 

4 

The  above  results  are  also  the  sine  and  cosine,  respec- 
tively, of  54°  (Art.  15). 

Otherwise  thus:  Let  x  =  36°;  then  2  x  =72°,  and  3  x  =108°. 


(Art.  29) 
2  sin  &•  cos  x  =  3  sin  #  —4  sin3  a?, 
2  cos  #==  3  —  4  sin2  a; 
=  4  cos2  x  —  1. 


Solving,  cos  0  =  ± 


But  36°  is  an  acute  angle,  and  therefore  its  cosine  is 
positive. 

...  oos  36'=^+! 
4 

59.  If  A  +  B  +  C  =  180°,  or  if  A,  B,  C  are  the  Angles 
of  a  Triangle,  prove  the  Following  Identities: 

ABC 

(1)  sin  A  +  sinB  +  sin  C  =  4cos  —  cos  —  cos  — 

222 

• 

ABC 

(2)  cos  A  -f  cosB  +  cos  C  =  1  +4  sin  —  sin  —  sin—  • 

222 

(3)  tan  A  -f-  tan  B  +  tan  C  =  tan  A  tan  B  tan  C. 


ANGLES   OF  A   TRIANGLE.  71 

We  have  A  +  B  +  C  =  180°. 
/.  sin(A+B)  =sinO,  and  sin^±I>  =  cos--  (Arts.  15  and  30) 


and 


.-.  sin  A  +  sinB  +  sin  C  =  2  cos  -  cos  - 

22 


J  AQLIL  —  UUO  

-  .      ^.rvirb.  *±uy 

.     (Art.  15) 

D  =  2sin-cos-  .     .     . 

2        2 

.     (Art.  49) 

9       «A  +  B           C 

.     (Art.  15) 

rt                          O 

^  —  2  cos  C  cos  A~B  -4-2< 

C       A+B 

=  2  cos  ^  (2  cos  £  cos  l^j  .     (Art.  45) 

2  y         2        2  J 

=  4cos|cos|cos|    ....     (1) 


Again,    cos  A  +  cos  B  =  2  cos    --    cos  A~B  .    (Art.  45) 


and  cosC  =  l-2sin25    ....     (Art.  49) 


.-.  cosA+ 


msinSin         .     .     (2) 


72  PLANE  TRIGONOMETRY. 

Again,     tan  (A  +  B)  =  —  tanC     .....     (Art.  30) 

=   tanA  +  tanB  (A      4  } 

1  -tan  A  tan  B 

• 

.-.  tanA  +  tanB  =  —  tanC  (1  —  tanAtanB). 
.-.  tanA  +  tanB+tanC  =  tanAtanB  tanC     ....     (3) 

NOTE.  —  The  student  will  observe  that  (1),  (2),  and  (3)  follow  directly  from 
Examples  1  and  2,  and  formula  (3),  respectively,  of  Art.  48,  by  putting 

A  +  B  +  C  =  180°. 
EXAMPLES. 

Prove  the  following  statements  if  A  +  B  +  C  =  180°  : 

1.  cos(A  +  B-C)  =  -cos2C. 

ABC 

2.  sinA  +  sinB  —  sinC  =  4sin—  sin  —  cos-- 

222 


3.  sin2A  +  sin2B  +  sin2C  =  4sinAsinB  sinC. 

4.  sin  2  A  +  sin  2  B  —  sin  2  C  =  4  sin  C  cos  A  cos  B. 

5.  tan  7  A  —  tan  4  A  —  tan  3  A  =  tan  7  A  tan  4  A  tan  3  A. 

6.  sin  A  —  sin  B  -f  sin  C  =  4  sin  —  cos  -  sin  —  • 

222 

7.  Cot  -  +  cot  ?  +  cot-  =  cot  ^  cot  §  cot?- 

8.  tan  A  —  cot  B  =  sec  A  cosec  B  cos  C. 

60.  Inverse  Trigonometric  Functions.  —  The  equation 
sin  6  —  x  means  that  0  is  the  angle  whose  sine  is  x  ;  this 
may  be  written  6=  sin"1^,  where  sin"1^  is  an  abbreviation 
for  the  angle  (or  arc)  whose  sine  is  x. 

So  the  symbols  cos"1^  tan"1^  and  sec"1!/,  are  read  "the 
angle  (or  arc)  whose  cosine  is  x"  "  the  angle  (or  arc)  whose 
tangent  is  #,"  and  "  the  angle  (or  arc)  whose  secant  is  y." 
These  angles  are  spoken  of  as  being  the  inverse  sine  of  x,  the 


INVERSE  TEIGONOMETEIC  FUNCTIONS.  73 

inverse  cosine  of  x,  the  inverse  tangent  of  x,  and  the  inverse 
secant  ofy,  respectively.  Such  expressions  are  called  inverse 
trigon  ometric  functions. 

NOTE.  —The  student  must  be  careful  to  notice  that  —  1  is  not  an  exponent,  sin'1  a; 

is  not  (sin  a:)'1,  which  =  — — 
sin  a 

Notice  also  that  sin"1  —  =  cos"1  -  is  not  an  identity,  but  is  true  only  for  the  par- 
ticular angle  60°. 

This  notation  is  only  analogous  to  the  use  of  exponents  in  multiplication,  where 
we  have  a-1tt  =  «°  =  l.  Thus,  cos"1  (cos x)  =  x,  and  sin  (sin'1 «)  =x;  that  is,  cos"1 
is  inverse  to  cos,  and  applied  to  it  annuls  it;  and  so  for  other  functions. 

The  French  method  of  writing  inverse  functions  is 
arc  sinx,  arc  cosx,  arc  tanx,  and  so  on. 


EXAMPLES. 

1.  Show  that  30°  is  one  value  of  sin"1  \. 

We  know  that  siii30°=  \.     .-.  30°  is  an  angle  whose  sine 
is  J;  or  30°=  sin-1  £. 

2.  Prove  that  tan-1  £  +  tan-1  \  =  45°. 

tan"1^-  is  one  of  the  angles  whose  tangent  is  J,  and  tan"1-^ 
is  one  of  the  angles  whose  tangent  is  J. 

Let  a  =  tan-1^-,     and     /8  =  tan~1^; 

then  tan  a  =  J     and     tan  (3  =  ^. 


Now        tan(«  +  /3)=  .     .     .     (Art.  47) 

1  —  tanatan/J 

-t±i_=l 

i-fxi 

But        tan  45°=  1,         .-.  a  +  £  =  45°; 
that  is,  taa-1  £  +  tan'1  J  =  45°. 

Therefore  45°  is  one  value  of  tan-1|-  +  tan-J|-. 


PLANE   TRIGONOMETRY. 

i-i^tau-'-^L 


3.   Prove  that  tan-1  a  +  tan-1?/ 


"1?/  =  B.       .•.  tan  B  =  y. 


Now 


—  xy 


l-xy       . 

Any  relations  which  have  been  established  among  the 
trigonometric  functions  may  be  expressed  by  means  of  the 
inverse  notation.  Thus,  we  know  that 


cos  x  =  Vl  —  sin2  a;. 


This  may  be  written          x  =  cos"1  Vl  —  sin2  a;    .     .     (1) 
Put          sino;  =  0;   then  a;  =  siii~10. 


Thus  (1)  becomes      sin-1  B  =  cos"1  Vl  -  0*. 

5.   By  Art.  49,  cos  26  =  2  cos2  0  -  1, 

which  may  be  written         26  =  cos"1  (2  cos2  0  —  1). 
Put   cosO  =  x.     .-.  2  cos-1  x=  cos-1  2^  —  1. 


6.   By  Art.  49,  sin  2  0  =  2  sin  0  cos  0, 

which  may  be  written         20  =  sin"1  (2  sin  0  cos  0)  . 


Put    sin  0  =  x.     .-.  2  sin-1  a  =  sin-1  (2  a  Vl- 


TABLE  OF  USEFUL  FORMULAE.                     75 
7.   Prove  sin-1a;  =  cos-1  Vl  —  x2  =  tan-1 — — 


8.       "  tan"1  x  =  sin-1  —  x       = 


cos' 


Vl  +  x2  Vl  +  x2 

9.       " 


1  —  or* 

10.  «     sin  (2  sin-1  a?)  =  2  a;  vT^ 

11.  « 


12.  «  cos-1-  +  2  sin-1  -  =  120°. 

13.  "  cot-1  3  +  cosec-1  V5  =  -• 

4 

14.  « 


61.  Table  of  Useful  Formulae.  —  The  following  is  a  list 
of  important  formulae  proved  in  this  chapter,  and  summed 
up  for  the  convenience  of  the  student  : 


.     .     (Art.  43) 

2.  cos  (a;  +  y)    =  cos  x  cos  y  —  sin  x  sin  y. 

3.  sin(a;  —  y)    =  sin  x  cos  y  —  cos  x  sin  y   .     .     (Art.  44) 

4.  cos  (x  —  y)    =  cos  x  cosy  +  sinx  siny. 

5.  2  sin  xcosy  =  sin  (a;  -f  y)  +  sin  (a;  —  y)     .     (Art.  45) 

6.  2coscc  siny  =  sin  (x  -f-  y)  —  sin  (a;  —  y). 

7.  2  cos  a;  cos  y  =  cos  (#  -f-2/)  +  cos  (x  —  y). 

8.  2sino;sin2/   =  cos  (x  —  y)  —  cos  (a;  +  y)  . 

9.  sinas-f  sin?/  =  2sin-j-(a;  +  y)  co^J(a;  —  y). 
10.  sinz  —  siny  = 


76  PLANE  TRIGONOMETRY. 

11.  coscc-j-  cos?/  =  2cos-J-(#  -|-  y)  cos^(x  —  y). 

12.  cosy  —  cos  x  =  2  sin  -J  (x  +  y)  siu^(x  —  y). 

13.  sin  a;  +  sin?/  =  tan^(gp  +  y) 
sin  a;  —  sin  y      tan  J  (a;  —  y  ) 


14.  tan  (x  +  ?/)    =  .  (Art.  47) 

l-tanztan?/ 

15.  tan(s-y)    = 


1  +  tan  x  tan  y 
16.    cot 


cot  x  -f-  coty 
17. 


COt  07  —  COt?/ 

18.   tan(a;±450)  =  tana;Tl       ....  (Art.  47) 

tan  x  ±  1 


19.  sin  (#-f  ?/)  sin  (x  —  y)  =  sin2  a;  —  sin2?/  =  cos2?/  —  cos2  a?. 

20.  cos  (x+  y)  cos  (a;  —  y)  =  cos2  a?  —  sin2?/  =  cos2?/  —  sin2  a?. 

21.  tan*  ±  tony  =  ?!°£[±*!. 

cos  a?  cos  ?/ 

22. 


sin  a;  sin?/ 
23.   sin2a?  =  2sina;cosa;=    2tana;  (Art.  49) 


24.   cos  2  a?  =  cos2  a;  —  sin2#=  1  —  2  sin2  a;  =  2  cos2  a  — 

_  1  —  tan2a; 
~  1  +  tan2  a;' 


1-|-  cos  2  x     2  cos2  a? 


26.  tan  2  a;  = 

27.  cot  2  a;  = 


EXAMPLES.  77 

2  tana 


1  —  tan2  a; 

COt2  a?  —  1 


2cot« 

28.  sin3x  =  3  sin  a; -4  sin3  a; (Art.  50) 

29.  cos  3  x  =  4  cos3  x  —  3  cos  x. 

3  tan  a;  —  tan3  a; 

30.  tan3a  =  -  — 

1  —  3  tan2  a; 

31.  sin2^   =1-<cosa? (Art.  51) 

9X          1  4-  COS  X 

32.  cos2-  =— ^-     — 

2  2 

33.  tan"1  a;  -\-  tan"1  y  =  tan"1         ^     ....     (Art.  60) 

EXAMPLES. 

1  2 

1.    If  sin  a  =  -,  and  sin  /?  =  -,  find  a  value  for  sin  («+/?), 

3  3 

and  sin  (a  —  ft).  A        V5  +  4V2.    V5-4V2T 


2.  If  cos  a  =  -,  and  cos  fi  =  — ,  find  a  value  for  sin  ( a  -f  /?) , 

o  41 

andcos(a  +  /8).  .        156     133 

!'   205'    205' 

o  o 

3.  If  cos  «  =  -,  and  cos  /?  =  -,  find  a  value  for  sin  (a  +  )8), 

4  5 

and  sin  («  -j8).  ^ng    2  V7  +  3  V21     2V7-3V21 

4.  If  sin  «  =  -,  and  sin/S  =  -,  find  a  value  for  sin  («  +  /?), 

andcos(«  +  /3).  ^Q    1     34. 

'    25 


78  PLANE   TRIGONOMETRY. 

5.  If  sin  a  =  .6,  and  sin  (3  =  — ,  find  a  value  for  sin  (a  —  /?), 
and  cos  («  +  /»).  16     33 

•   65'    66' 

6.  If  sin«= — -,  and  sin  5  =  — =,  show  that  one  value 

V5  Vio 

of  a  +  0  is  45°. 

7.  Prove  cos  6  4-  cos  3  0   =2  cos  2  0  cos  0. 

8.  "      2cosacos/8       =  cos  (a  —  (3)  +  cos  (a  +  /?). 

9.  "      2sin30cos50  =sin80-sin20. 

10.  "      2cos|0cos-     =cos^-f  cos20. 

11.  "      sin 40  sin  0        =  £(cos30  -  cos  50). 

12.  "      2  cos  10°  sin  50°  =  sin  60°  +  sin  40°. 

13.  Simplify  2 cos  2  9  cos  6  —  2  sin  4  0  sin  0. 

^Ins.  2  cos  30  cos  20. 

50        0  00        S0 

14.  Simplify  sin — cos sin  —  cos —      —cos 40 sin 20. 

22  22 

Prove  the  following  statements  : 

15.  cos3«  —  cos7a==  2 sin  5 «  sin  2 a. 

16.  sin  60°+  sin  20°=  2  sin  40°  cos  20°. 

17.  sin30-fsin50  =2sin40cos0. 

18.  sin70-sino0  =2cos60sin0. 

19.  cos50  +  cos90  =  2cos70cos20. 

20.  Sin2*  +  sin*    =tan^- 
cos0  +  cos20  2 

21.  cos  (60°  +  A)  +  cos  (60°  -  A)  =  cos  A. 


EXAMPLES,  79 

22.  cos  (45°+  A)  +  cos  (45°-  A)  =  V2cos  A. 

23.  sin  (45°+  A)  -sin  (45°-  A)  =  V2sinA. 

24.  cos20  +  cos40  =  2cos30cos0. 

25.  cos40  —  cos60  =  2sin50sin0. 

26.  cos  0  +  cos  3  6  +  cos  5  0  +  cos  7  0  =  4  cos  0  cos  2  0  cos  4  0. 

27. 


sma  cos/3 

28.  ooU-tan/8   =cos<"!  +  ^ 

sin  a  cos  /? 

29.  sinA 


V2 

30.  V2  sin  (A  +  45°)  =  sin  A  -f  cos  A. 

31.  cos  (A  +  45°)  +  sin  (A-  45°)  =  0. 

32.  tan  (*-*)+  ten*  = 
1  —  tan  (0  —  <£)  tan  <f> 


33.  = 


34.  cos  (6  +  <£)  -  sin  (0  -  <£)  =  2sin    -  -  0}cos   -  - 

V4        / 

35.  sin  ?i0  cos  0  +  cos  nO  sin  0  =  sin  (n  -f  1  )  0. 

36.  c 


l-cot0 

37.  tan  ($-*]  +  cot  (o  +  -    =  0. 

V 

38.  cotf0- 

V 

39.  tan  (n  +  1)0-  tan  n^   =  tap 


80  PLANE   TRIGONOMETRY. 

40.  If  tanx  =  1,  and  tany  =  —  -,  prove  that 

V  O 

tan  (a?  +  y)  =  2  +  VS. 

41.  If  tana  =  —  —  —  ,  and  tan/?  =  -  ,  prove  that 

m+1  2m  +1 

tan(«4-/3)  =  l. 

42.  If  tan  a  =  m}  and  tan  ft  =  w,  prove  that 

1~mn 


cos 


43.  If  tan0  =  (a  -f-  1),  and  tan  <£  =  (a  —  1),  prove  that 

2  cot  (0  -</>)  =  a2. 

Prove  the  following  statements  : 

44.  cos  OK  —  2/4-  z)  =  cosx  cosy  cos z  -\-  cosx  sin?/  sinz 

—  sin  x  cos  y  sin  2  -f  sin  x  sin  ?/  cos  2. 

45.  sin  (x  —  y  —  z)  =  sin  x  +  sin?/  +  sins; 

4-  4sin  J(x—  y)  sin  J(x— 2)  sini(?/-f  2 

46.  sin  (B  -f-  y  —  z)  +  sin  (#  -f-  z  —  y)  -f  sin  (y  -\-  z  —  x) 

=  sin  (a;  -}-  y  -f-  z)  4-  4  sin  a;  sin  ?/  sin  z. 

47.  sin2#  +  sin2?/-f-  sin 2z—  sin2  (x  +  y  -f«) 

=  4  sin  (ic  +  y)  sin  (y  -\-  z)  sin  (z-\-  x). 

48.  cos  2  a;  4-  cos  2y  -{-  cos  2  2  4-  cos  2  (x  +  y  -\-z) 

=  4 cos  (a;  4-  ?/)  cos  (y  4-  z)  cos  («  4-  a). 

49.  cos  (x  +  i/  —  2)  4  cos  (y  +  2  —  0,-)  4-  cos  (2  +  a;  —  y) 

4-  cos  (a;  4-  y  +  2)  =  4  cos  x  cos  ?/  cos  z. 


EXAMPLES.  81 

50.  sin2 a;  -f-  sin2?/  -f  sin2z  -f-  sin2  (x  -f  y  -f  z) 

=  2jl  —  cos  (#  +  ?/)  cos  (y+z)  cos  (z-f-a)  \. 

51.  cos2 a;  -f  cos2?/  +  cos2z  +  cos2  (aj  +  y  —  z) 

=  2jl  -f-  cos  (a?-f  ?/)  cos  (x—z)  cos  (y—z)  \. 

52.  cosa;  sin  (y  —  z)  -f-  cos?/  sin(z  —  a;)  +  cosz  sin(a;—  ?/)  =  0. 

53.  sin  x  sin  (?/  —  z)  +  sin  ?/  sin  (z — a;)  -f  sin  z  sin  (x  —  y)  =  0. 

54.  cos  (a;  -f-  y)  cos  (#  —  ?/)-}-  sin  (y  +  z)  sin  (?/  —  z) 

—  cos  (a?  -f  2)  cos  (a;  —  z)  =  0. 

O  2/1 

55. 


sec20 

56.  cos2  0(1-  tan2  0)  =  cos  2  0. 

57.  cot20  =  cot^^l. 

2cot0 

58.  sec20=" 


cot20-l 

59.  ( sin -  +  cos- )  =  1  +  sin  ft 
V     2         2y 

60.  (sin  -  —  cos  -  )  =  1  —  sin  0. 
\     2          2y 


61. 

2 


62        cos  2  0         1  —  tan  0 


63. 


l-tan| 


82  PLANE  TRIGONOMETRY. 

gj     1  +  sin  x  -f-  cos  x  _       x 
1  +  sin  x  —  cos  a;  2 

c~     cos3 x  -h  sin3 x 
bo. 


cos  x  -f  sin  x  2 


~~  —  sin3x          2  +  sin2a; 

DO.       -  -  ;  -  =  -  • 

cos#  —  smas  2 

67.    cos40-sin4<9      =  cos  20. 

aQ     sin  30     cos  30 
bo.    ---    =  £. 
smO        cosO 


69.    cosSf     sinSS   =2cot2ft 
sin  ^        cos  0 


70. 


sin  20 
71. 


sin        cos 


72.  tan  (45°  +  a?)  -  tan  (45°  -  a?)  =  2  tan  2  a?. 

73.  tan  (45°  -  x)  +  cot  (45°  -  a;)  =  2  sec  2  a. 

74.  tanW+x)-!^^ 
tan2  (45°+  a)  +  1 

-rf          COS   (iC  +  45°) 

75.  .   __v  —  Z  -  i  =  sec  2  ic  —  tan  2  ic. 
cos  (a?  -  45°) 

76.  tans=      sin  a;  +  sin  2  a;    t 

1  +  cos  a;  +  cos  2  x 

77. 


1  —  cos  x  -f-  cos  2  a; 


78.    Cos3a; 
cos  a; 


EXAMPLES.  83 

QOIYI/Y*  01  n  Q  />• 

79. 


cos  3  x  +  3  cos  x 

cot8  a?  —  3  cot  a; 
80.    cot3#  =  — 

3  cot2  a;  — 1 

Q1     1  — cos3ir 

51.    — 

1  —  cos  a; 

82. 


cos  x  —  sin  x 

go     cos 2 x  -f-  cos  12 x     cos  7  a  —  cos 3 x     2sin4# 
cos  6  x  -f  cos  8  x        cos£  —  cos  3  a?        sin2x 

84.  sin2a;  sm2y  =  sin2(«  +  y)  —  sin2(o;  —  y). 

85.  tan50°H-cot50°=2seclO°. 

86.  sin  3  a  =  4  sin  a  sin  (60°  +  a)  sin  (60°  -  x) . 

87.  cot --tan- =  2. 

8  8 

88. 


89. 


90.    (3sin0  -  4sin30)2+  (4cos30  - 

91  sin 2 6>  cos (9  =t^e 

(lH-cos2(9)(l+cos0)  2 

92.    If  tan  0  =  -,  and  tan  «#>  =  —,  prove  tan  (2  0  +  <£)  =  i- 

95.    Prove  that  tan-    and   cot-    are   the   roots   of  the 

2  2 

equation 


84  PLANE  TRIGONOMETRY. 


94.    If  tan0  =  -,  prove  that 
a 


6         la  -b        2cos0 


—  b      \  a  +  b      Vcos  2  0 

95.    Find  the  values  of  (1)  sin  9°,  (2)  cos  9°,  (3)  sin  81°, 
(4)  cos  189°,  (5)  tan202i°,  (6)  tan97|°. 


Ans.   (1)  i  ( V3  +  V5  -  V5  -  V5), 


(2)  1(^3  +  V5+-  V5-  V5), 

(3)  sin  81°=  cos  9°, 

(4)  cos  189°= -cos 9°, 

(5)  V2  -  1, 

(6)  - 
96.    If  A  =  200°,  prove  that 


(1)  2sin^  =  +  Vl-f-sinA  +  Vl-sjnA. 


2  tan  A 

97.  If  A  lies  between  270°  and  360°,  prove  that 

(1)  2sin-  =  +  Vl  -  sin  A  -  Vl  +  sinA. 

2 

(2)  tan  -  =  —  cot  A  +  cosec  A. 

ft 

98.  If  A  lies  between  450°  and  630°,  prove  that 


.    A 


(1)  2  sin—  =  —  Vl  +  sin  A  —  Vl  —sin  A. 


(2)  2cos— =  —  Vl  -fsinA  +  Vl  —  sin  A. 


EXAMPLES.  85 

Prove  the  following  statements,  A,  B,  C  being  the  angles 
of  a  triangle. 

99. 
100. 


sin  A  +  sin  B  2  2 

sin3B  — sin3C  _tan3A 
cos3C-cos3B~ 


101.    sin  -  cos  -  +  sin  2  cos  ?+  sin-  cos  - 

222222 


102.  cos2^  +  cos2|-  cos2^  =  2  cos-  cos?  sin  5. 

103.  sin  A  cos  A  —  sin  B  cos  B  -f  sm  C  cos  C 

=  2  cos  A  sin  B  cos  C. 

104.  cos  2  A  +  cos  2  B  +  cos2C  =  —  1  —  4  cos  A  cos  B  cos  C. 

105.  sin2  A  —  sin2  B  +  sin2  C  *=  2  sin  A  cos  B  sin  C. 

106.  tan  ?  tan  -  +  tan  -  tan  -  +  tan  -  tan  -  =  1. 

2         2  2        2  2        2 

Prove  the  following  statements  when  we  take  for  sin"1, 
cos"1,  etc.,  their  least  positive  value. 

107.  sin-1  -  =  008-^  =  cot"1  V3. 

2i  2i 

108.  2tan-'(cos2e)  = 
109. 


86  PLANE  TRIGONOMETRY. 

111.  tan-1  V5  (2  -  V3)  -  cof1  V5  (2  +  V3)  =  coir1  VB. 

112.  sec-1V3 


1  4-  X2 

113.    2  cot"1  a;  =  cosec"1        — 


115.    Bin-l=: 

V5 


116. 


117.  If  6  =  sin-1-,  and  <£  =  cos-1?  then  0  +  4  =  90°. 

5  5 

1    _  ^,2 

118.  Prove  that  cos  (2  tan-1  a)  =  —   '—> 

1  -f-  x" 

119         "         "    tan-1^  ^^  -h  cosec-1  VlO  =  ^- 
2  4 

2  5  33 

120.        "         "     2tan-V  —  cosec-1-  =  sin-1—  • 

O  O  OO 


121. 

122.  "         "     sin-1  (cos  x)  +  cos"1  (sin  y)  +  cc  +  y  =  TT. 

1  12 

123.  "         «     tan-1— h  tan-1-     -  +  tan"1  -5  =  »TT. 

1  _i_  x  1  —  a;  or 


124.  "         "     tan-1—   -  +  tan-1  — -  —  =  nir  +  - 

125.  "         "     sin-1  x  —  sin-1  y 


=  cos-1  (xy  ±  VI  -  w2  -  2/2  4-  *T 


NATURE  AND    USE  OF  LOGARITHMS.  87 


CHAPTER   IV. 

LOGARITHMS   AND  LOGARITHMIC   TABLES,  —  TRIGO- 
NOMETRIC TABLES, 

62.  Nature  and  Use  of  Logarithms.  —  The  numerical  cal- 
culations   which    occur   in    Trigonometry   are    very   much 
abbreviated  by  the  aid  of  logarithms ;  and  thus  it  is  neces- 
sary to  explain  the  nature  and  use  of  logarithms,  and  the 
manner  of  calculating  them. 

The  logarithm  of  a  number  to  a  given  base  is  the  exponent 
of  the  power  to  which  the  base  must  be  raised  to  give  the 
number. 

Thus,  if  ax  =  m,  x  is  called  the  "  logarithm  of  m  to  the 
base  a,"  and  is  usually  written  o?  =  logam,  the  base  being 
put  as  a  suffix.* 

The  relation  between  the  base,  logarithm,  and  number  is 
expressed  by  the  equation, 

(base)log  =  number. 

Thus,  if  the  base  of  a  system  of  logarithms  is  2,  then  3 
is  the  logarithm  of  the  number  8,  because  23  =  8. 

If  the  base  be  5,  then  3  is  the  logarithm  of  125,  because 
53  =  125. 

63.  Properties  of  Logarithms.  —  The  use  of  logarithms 
depends  on  the  following  properties  which  are  true  for  all 
logarithms,  whatever  may  be  the  base. 

*From  the  definition  it  follows  that  (1)  loga  a*  =  x,  and  conversely  (2)  alo&a"»  =  m. 
Taking  the  logarithms  of  both  sides  of  the  equation  ax  =  m,  we  have  loga  ax  =  x=\og  m. 
Conversely,  taking  the  exponentials  of  both  sides  of  x  =  loga  m  to  base  a,  we  have 
ax  =  olo?am  =  m.  ax  =  m  and  x  =  \ognm  are  thus  seen  to  be  equivalent,  and  to 
express  the  same  relation  between  a  number,  in,  and  its  logarithm,  x,  to  base  a. 


88  PLANE  TRIGONOMETRY. 

(1)  The  logarithm  of  1  is  zero. 

For  a°  =  1,  whatever  a  may  be ;  therefore  log  1  =  0. 

(2)  The  logarithm  of  the  base  of  any  system  is  unity. 
For  al=i  a,  whatever  a  may  be  ;  therefore  logaa  =  1. 

(3)  The  logarithm  of  zero  in  any  system  whose  base  is 
greater  than  1  is  minus  infinity. 

For  or00  =  —  =  -  -  =  0  :  therefore  log  0  =  —  cc . 
a00      co 

(4)  The  logarithm  of  a  product  is  equal  to  the  sum  of  the 
logarithms  of  its  factors. 

For  let  x  =  logaw,  and  y  —  loga?i. 

.-.  m  =  ax,         and  n  =  ay. 
.•.  mn  =  ax+y. 

.•.  logamn  =  x  +  y  =  logam  +  logan. 
Similarly,          logamnp  =  logam  +  logaw  +  \°gaP> 
and  so  on  for  any  number  of  factors. 

Thus,  log  60  =  log  (3  x  4  x  5), 

=  log  3  + log  4  + log  5. 

(5)  The  logarithm  of  a  quotient  is  equal  to  the  logarithm 
of  the  dividend  minus  the  logarithm  of  the  divisor. 

For  let  #  =  logara,  and  y  =  logaw. 

.  •.  m  =  ax,         and  n  =  a*. 


.-.  loga—  =  x  —  y  =  logam  —  logan. 
Thus,  log—  =  log  17  -  log 5. 


PROPERTIES   OF  LOGARITHMS.  89 

(6)  The  logarithm  of  any  power  of  a  number  is  equal  to 
the  logarithm  of  the  number  multiplied  by  the  exponent  of  the 
poiver. 

For  let  aj  =  logam.     .-.  m  ==  a*. 

.-.  mp  =  apx. 
.  :  loga  mp  =  px  =  p  loga  m. 

(7)  The  logarithm  of  any  root  of  a  number  is  equal  to  the 
logarithm  of  the  number  divided  by  the  index  of  the  root. 

For  let  x  =  logam.     .-.  m  =  a*. 

1  X 

.-.  mr  =  ar. 

1       x     1 
.-.  log(mr)  =  -  =  -logam. 

It  follows  from  these  propositions  that  by  means  of 
logarithms,  the  operations  of  multiplication  and  division  are 
changed  into  those  of  addition  and  subtraction;  and  the 
operations  of  involution  and  evolution  are  changed  into  those 
of  multiplication  and  division. 

1.  Suppose,  for  instance,  it  is  required  to  find  the  product 
of  246  and  357;  we  add  the  logarithms  of  the  factors,  and 
the  sum  is  the  logarithm  of  the  product :  thus, 

Iog10  246  =  2.39093 

Iog10357  =  2.55267 

4.94360 

which  is  the  logarithm  of  87822,  the  product  required. 

2.  If  we  are  required  to  divide  371.49  by  52.376,  we  pro- 
ceed thus : 

logw  371.49  =  2.56995 
logM  52.376  =  1.71913 

0.85082 
which  is  the  logarithm  of  7.092752,  the  quotient  required. 


90  PLANE  TRIGONOMETRY. 

3.    If  we  have  to  find  the  fourth  power  of  13,  we  proceed 
thus: 

Iog10 13  =  1.11394 
4 


4.45576 
which  is  the  logarithm  of  28561,  the  number  required. 

4.  If  we  are  to  find  the  fifth  root  of  16807,  we  proceed 
thus : 

5)4.22549  =  Iog10 16807, 

0.845098 
which  is  the  logarithm  of  7,  the  root  required. 

5.  Given  Iog102  =  0.30103 ;  find  Iog10128,  Iog10512. 

Ans.  2.10721,  2.70927. 

6.  Given  Iog103  =  0.47712 ;  find  Iog1081,  Iog102187. 

Ans.  1.90849,  3.33985. 

7.  Givenlog103;   find  log]0  -tys*.  0.28627. 

8.  Find  the  logarithms  to  the  base  a  of 

o        -V>        4/~       3/— r,        -5 

a3,  a  3 ,  y  a,  y  a2,  a  2. 

9.  Find  the  logarithms  to  the  base  2  of  8,  64,  £,  .125, 
.015625,  -v/64.  Ans.  3,  6,   -1,   -3,   -6,  2. 

10.  Find   the   logarithms    to   base   4   of    8,    A/16,    ^.5? 
AV.015625.  Ans.  f,  f,   -  J,   -  1. 

Express  the  following  logarithms  in  terms  of  log  a,  log  6, 
and  logc: 

11.  logVO'^c)6.  Ans.  61og«  +  91og6  +  31ogc. 

floga  +  |  log  6  -f  !logc. 


SYSTEMS   OF  LOGARITHMS.  91 

64.  Common  System  of  Logarithms.  —  There  are  two 
systems  of  logarithms  in  use,  viz.,  the  Naperian*  system 
and  the  common  system. 

The  Naperian  system  is  used  for  purely  theoretic  investi- 
gations; its  base  is  e  =  2.7182818. 

The  common  system  f  of  logarithms  is  the  system  that 
is  used  in  all  practical  calculations  ;  its  base  is  10. 

By  a  system  of  logarithms  to  the  base  10,  is  meant  a  suc- 
cession of  values  of  x  which  satisfy  the  equation 

m  =  10*, 

for  all  positive  values  of  m,  integral  or  fractional.  Thus,  if 
we  suppose  m  to  assume  in  succession  every  value  from  0 
to  oo,  the  corresponding  values  of  x  will  form  a  system  of 
logarithms,  to  the  base  10. 

Such  a  system  is  formed  by  means  of  the  series  of  loga- 
rithms of  the  natural  numbers  from  1  to  100000,  which  con- 
stitute the  logarithms  registered  in  our  ordinary  tables. 

Now  10°  =1,  .-.  logl        =0; 

10T=:10,  .-.  log  10      =1; 

102=100,  .-.  log  100    =2; 

103=1000,  .-.  log  1000  =  3. 

and  so  on. 

Also,  10^=  TV      =  .1,         .-.  log.l       =-1; 

10"2=yU    =-01>       -•-  log.Ol     =-2; 
10-3=  T¥VTF  -  -001,     .-.  log.OOl  =  -  3. 


and  so  on. 

Hence,    in   the   common   system,  the  logarithm   of   any 
number  between 

1  and  10  is  some  number  between  0  and  1  ;    i.e.,  0  + 
a  decimal  ; 

*  So  called  from  its  inventor,  Baron  Napier,  a  Scotch  mathematician. 
f  First  introduced  in  1G15  by  firiygs,  a  contemporary  of  Napier. 


92  PLANE   TRIGONOMETRY. 

10  and  100  is  some  number  between  1  and  2 ;  i.e.,  1  -f- 
a  decimal ; 

100  and  1000  is  some  number  between  2  and  3 ;  i.e.,  2  -f 
a  decimal ; 

1  and  .1  is  some  number  between  0  and  —  1 ;  i.e.,  —  1  + 
a  decimal ; 

.1  and  .01  is  some  number  between   —  1  and  —  2 ;  i.e., 

—  2  +  a  decimal ; 

.01  and  .001  is  some  number  between  —  2  and  —  3 ;  i.e., 

—  3  +  a  decimal ; 

and  so  on. 

It  thus  appears  that 

(1)  The  (common)  logarithm  of  any  number  greater  than 
1  is  positive. 

(2)  The  logarithm  of  any  positive  number  less  than  1  is 
negative. 

(3)  In  general,  the  common  logarithm  of  a  number  con- 
sists of  two  parts,  an  integral  part  and  a  decimal  part. 

The  integral  part  of  a  logarithm  is  called  the  characteristic 
of  the  logarithm,  and  may  be  either  positive  or  negative. 

The  decimal  part  of  a  logarithm  is  called  the  mantissa 
of  the  logarithm,  and  is  always  kept  positive. 

NOTE.  — It  is  convenient  to  keep  the  decimal  part  of  the  logarithms  always  posi- 
tive, in  order  that  numbers  consisting  of  the  same  digits  in  the  same  order  may 
correspond  to  the  same  mantissa. 

It  is  evident  from  the  above  examples  that  the  character- 
istic of  a  logarithm  can  always  be  obtained  by  the  following 
rule : 

RULE.  —  The  characteristic  of  the  logarithm  of  a  number 
greater  than  unity  is  one  less  than  the  number  of  digits  in 
the  whole  number. 

The  characteristic  of  the  logarithm  of  a  number  less  than 
unity  is  negative,  and  is  one  more  than  the  number  of  ciphers 
immediately  after  the  decimal  point. 


RULES  FOR    THE  CHARACTERISTIC.  93 

Thus,  the  characteristics  of  the  logarithms  of  1234,  123.4, 
1.234,  .1234,  .00001234,  12340,  are  respectively,  3,  2,  0,  -  1, 
-5,  4. 

NOTE.  —  When  the  characteristic  is  negative,  the  minus  sign  is  written  over  it  to 
indicate  that  the  characteristic  alone  is  negative,  the  mantissa  being  always  positive. 

Write  down  the  characteristics  of  the  common  logarithms 
of  the  following  numbers  : 

1.  17601,  361.1,  4.01,  723000,  29.  Ans.  4,  2,  0,  5,  1. 

2.  .04,  .0000612,  .7963,  .001201,  .1. 

Ans.   -2,  -5,  -1,  -3,  -1. 

3.  How  many  digits  are  there  in  the  integral  part  of  the 
numbers  whose  common  logarithms  are  respectively  3.461, 
0.30203,  5.47123,  2.67101  ? 

4.  Given  log  2  =  0.30103;  find  the  number  of  digits  in  the 
integral  part  of  810,  212,  1620,  2100.  Ans.  10,  4,  25,  31. 

65.  Comparison  of  Two  Systems  of  Logarithms.  —  Given 
the  logarithm  of  a  number  to  base  a  ;  to  find  the  logarithm 
of  the  same  number  to  base  b. 

Let  ra  be  any  number  whose  logarithm  to  base  b  is 
required. 

Let  o;  =  log6m;  then  b*  =  m. 

logam;  or  xlogab  =  logam. 


Hence,  to  transform  the  logarithm  of  a  number  from 

base  a  to  base  b,  we  multiply  it  by  -- 

log.  b 


94  PLANE  TRIGONOMETRY. 


This  constant  multiplier  is  called  the  modulus  of 

log.  & 

the  system  of  which  the  base  is  b  with  reference  to  the  system 
of  which  the  base  is  a. 

If,  then,  a  list  of  logarithms  to  some  base  e  can  be  made, 
we  can  deduce  from  it  a  list  of  common  logarithms  by  mul- 
tiplying each  logarithm  in  the  given  list  by  the  modulus 

of  the  common  system • 

logelO 

Putting  a  for  m  in  (1),  we  have 

log,  a  =  ^  =  JL,  by  (2)  of  Art.  63. 
loga&      Iog06 

.-.  Iog6a  x  loga&  =  l. 


EXAMPLES. 

1.  Show  how  to  transform  logarithms  with  base  5  to 
logarithms  with  base  125. 

Let  m  be  any  number,  and  let  x  be  its  logarithm  to  base 
125. 

Then  m  =  125*  =  (53)*  =  5to.     .'.  3z  =  log5m. 


Thus,  the  logarithm  of  any  number  to  base  5,  divided  by 
3  (i.e.,  by  Iog5125),  is  the  logarithm  of  the  same  number  to 
the  base  125. 

Otherwise  by  the  rule  given  in  (1).     Thus, 


Show  how  to  transform 

2.   Logarithms  with  base  2  to  logarithms  with  base  8. 

Ans.  Divide  each  logarithm  by  3. 


TABLES   OF  LOGARITHMS.  95 

3.  Logarithms  with  base  9  to  logarithms  with  base  3. 

Am.  Multiply  each  logarithm  by  2. 

4.  Find  Iog2  8,  Iog5l,  Iog82,  Iog7l,  Iog32128. 

Ans.  3,  0,  |,  0,  £ 

66.  Tables  of  Logarithms.  —  The  common  logarithms  of 
all  integers  from  1  to  100000  have  been  found  and  registered 
in  tables,  which  are  therefore  called  tabular  logarithms.  In 
most  tables  they  are  given  to  six  places  of  decimals,  though 
they  may  be  calculated  to  various  degrees  of  approximation, 
such  as  five,  six,  seven,  or  a  higher  number  of  decimal  places. 
Tables  of  logarithms  to  seven  places  of  decimals  are  in 
common  use  for  astronomical  and  mathematical  calculations. 
The  common  system  to  base  10  is  the  one  in  practical  use, 
and  it  has  two  great  advantages  : 

(1)  From  the  rule  (Art.  64)  the  characteristics  can  be 
written  down  at  once,  so  that  only  the  mantissas  have  to  be 
given  in  the  tables. 

(2)  The  mantissas  are  the  same  for  the  logarithms  of  all 
numbers  which  have  the  same  significant  digits,  in  the  same 
order,  so  that  it  is  sufficient  to  tabulate  the  mantissas  of  the 
logarithms  of  integers. 

For,  since  altering  the  position  of  the  decimal  point  with- 
out changing  the  sequence  of  figures  merely  multiplies  or 
divides  the  number  by  an  integral  power  of  10,  it  follows 
that  its  logarithm  will  be  increased  or  diminished  by  an 
integer;  i.e.,  that  the  mantissa  of  the  logarithm  remains 
unaltered. 

In  General.  —  If  N  be  any  number,  and  p  and  q  any 
integers,  it  follows  that  N  X  10P  and  N  -5-  10*  are  numbers 
whose  significant  digits  are  the  same  as  those  of  K 


Then    log  (N  x  10P)  =  logN  +ploglO  =  logN  +p.     (1) 
Also,      log  (N  -s-  10')  =  log  N  -  q  log  10  =  log  N  -  q.      (2) 


96  PLANE   TRIGONOMETRY. 

In  (1)  the  logarithm  of  N  is  increased  by  an  integer,  and 
in  (2)  it  is  diminished  by  an  integer. 

That  is,  the  same  mantissa  serves  for  the  logarithms  of 
all  numbers,  whether  greater  or  less  than  unity,  which  have 
the  same  significant  digits,  and  differ  only  in  the  position 
of  the  decimal  point. 

This  will  perhaps  be  better  understood  if  we  take  a 
particular  case. 

From  a  table  of  logarithms  we  find  the  mantissa  of  the 
logarithm  of  787  to  be  895975  ;  therefore,  prefixing  the  char- 
acteristic with  its  appropriate  sign  according  to  the  rule, 
we  have 

log  787     =2.895975. 

Now  log  7.87     =  log  ~  =  log  787  -  2 

100 

=  0.895975. 
Also,  log  .0787  =  log  =  log  787  -  4 


=  2.895975. 

Also,  log  78700  =  log  (787  x  100)  =.  log  787  +  2 

=  4.895975. 

NOTE  1.  —  We  do  not  write  logto  787  ;  for  so  long  as  we  are  treating  of  logarithms 
to  the  particular  base  10,  we  may  omit  the  suffix. 

NOTE  2.  —  Sometimes  in  working  with  negative  logarithms,  an  arithmetic  artifice 
will  be  necessary  to  make  the  mantissa  positive.  For  example,  a  result  such  as 
—  2.69897,  in  which  the  whole  expression  is  negative,  may  be  transformed  by  sub- 
tracting 1  from  the  characteristic,  and  adding  1  to  the  mantissa.  Thus, 

-  2.69897  =  —  3  +  (1  -  .69897)  =  3.30103. 

NOTE  3.  —  When  the  characteristic  of  a  logarithm  is  negative,  it  is  often,  espe- 
cially in  Astronomy  and  Geodesy,  for  convenience,  made  positive  by  the  addition 
of  10,  which  can  lead  to  no  error,  if  we  are  careful  to  subtract  10. 

Thus,  instead  of  the  logarithm  3~.60S582,  we  may  write  7.603582  —  10. 

In  calculations  with  negative  characteristics  we  follow 
the  rules  of  Algebra. 


EXAMPLES.  97 


EXAMPLES. 

1.   Add  together  2.2143 

1.3142 

5.9068 


7.4353  Ans. 


2.   From  3.24569 

take  5.62493 


1.62076 

the  1  carried  from  the  last  subtraction  in  decimal  places 
changes  —  5  into  —  4,  and  then  —  4  subtracted  from  —  3 
gives  1  as  a  result. 

3.  Multiply  2.1528  by  7. 

2.1528 

7 

13.0696 

the  1  carried  from  the  last  multiplication  of  the  decimal 
places  being  added  to  —  14,  and  thus  -giving  —  13  as  a  result. 

NOTE  4.  —  When  a  logarithm  with  negative  characteristic  has  to  be  divided  by  a 
number  which  is  not  an  exact  divisor  of  the  characteristic,  we  proceed  as  follows  in 
order  to  keep  the  characteristic  integral.  Increase  the  characteristic  numerically  by 
a  number  which  will  make  it  exactly  divisible,  and  prefix  an  equal  positive  number 
to  the  mantissa. 

4.  Divide  3.7268  by  5. 

Increase  the  negative  characteristic  so  that  it  may  be 
exactly  divisible  by  5  ;  thus 

3.7268      5  +  2.7268 


Given  that  Iog2=.30103,  Iog3«=.47712,  and  Iog7=.84510; 
find  the  values  of 

5.   log  6,  log  42,  log  16.          Ans.  .77815,  1.62325,  1.20412. 


98  PLANE  TRIGONOMETRY. 

6.  log 49,  log 36,  log  63.     Am.  1.69020,  1.55630,  1.79934. 

7.  log 200,  log  600,  log  70.          2.30103,2.77815,1.84510. 

8.  log 60,  log. 03,  log  1.05,  log. 0000432. 

NOTE.  —  The  logarithm  of  5  and  its  powers  can  always  be  obtained  from  log 2. 

Ana.  1.77815,  2.47712,  .02119,  5.63548. 

9.  Given  Iog2=.30103;  find  log  128,  log  125,  and  log  2500. 

Ans.  2.10721,  2.09691,  3.39794. 

Given  the  logarithms  of  2,  3,  and  7,  as  above ;  find  the 
logarithms  of  the  following  : 

10.  20736,  432,  98,  686,  1.728,  .336. 

Ans.  4.31672,  2.63548,  1.99122,  2.83632,  .23754,1.52634. 

11.  V.~2,  (.03)*,  (.0021)*,  (.098)3,  (.00042)5,  (.0336) *. 
Ans.  1.65052,  1.61928, 1.46444,  4.97368,  17.11625, 1.26317. 

67.  Use  of  Tables  of  Logarithms  *  of  Numbers.  —  In  our 

explanations  of  the  use  of  tables  of  common  logarithms  we 
shall  use  tables  of  seven,  places  of  decimals. f  These  tables 
are  arranged  so  as  to  give  the  mantissse  of  the  logarithms 
of  the  natural  members  from  1  to  100000 ;  i.e.,  of  numbers 
containing  from  one  to  five  digits. 

A  table  of  logarithms  of  numbers  correct  to  seven  deci- 
mal places  is  exact  for  all  the  practical  purposes  of  Astron- 
omy and  Geodesy.  For  an  actual  measurement  of  any  kind 
must  be  made  with  the  greatest  care,  with  the  most  accurate 
instruments,  by  the  most  skilful  observers,  if  it  is  to  attain 
to  anything  like  the  accuracy  represented  by  seven  signifi- 
cant figures. 

*  The  methods  by  which  these  tables  are  formed  will  be  given  in  Chap.  VIII. 

t  The  student  should  here  provide  himself  with  logarithmic  and  trigonometric 
tables  of  seven  decimal  places.  The  most  convenient  seven-figure  tables  used  in  this 
country  are  Stanley's,  Vega's,  Bruhns',  etc.  In  the  appendix  to  the  Elementary 
Trigonometry  are  given  five-figured  tables,  which  are  sufficiently  near  for  most  prac- 
tical applications. 


USE  OF  LOGARITHMIC   TABLES.  99 

If  the  measure  of  any  length  is  known  accurately  to  seven  figures, 
it  is  practically  exact ;  i.e.,  it  is  known  to  within  the  limits  of  obser- 
vation. 

If  the  measure  of  any  angle  is  known  to  within  the  tenth  part  of  a 
second,  the  greatest  accuracy  possible,  at  present,  in  the  measurement 
of  angles  is  reached.  The  tenth  part  of  a  second  is  about  the  two- 
millionth  part  of  a  radian.  This  degree  of  accuracy  is  attainable  only 
with  the  largest  and  best  instruments,  and  under  the  most  favorable 
conditions. 

On  page  101  is  a  specimen  page  of  Logarithmic  Tables. 
It  consists  of  the  mantissse  of  the  logarithms,  correct  to 
seven  places  of  decimals,  of  all  numbers  between  62500  and 
63009.  The  figures  of  the  number  are  those  in  the  left 
column  headed  N,  followed  by  one  in  larger  type  at  the  top 
of  the  page.  The  first  three  figures  of  the  mantissas  795, 
796,  797  •••,  and  the  remaining  four  are  in  the  same  hori- 
zontal line  with  the  first  four  figures  of  the  number,  and  in 
the  vertical  column  under  the  last. 

Logarithms  are  in  general  incommensurable  numbers. 
Their  values  can  therefore  only  be  given  approximately. 
Throughout  all  approximate  calculations  it  is  usual  to  take 
for  the  last  figure  which  we  retain,  that  figure  which  gives 
the  nearest  approach  to  the  true  value.  When  only  a  cer- 
tain number  of  decimal  places  is  required,  the  general  rule 
is  this :  Strike  out  the  rest  of  the  figures,  and  increase  the  last 
figure  retained  by  1  if  the  first  figure  struck  off  is  5  or  greater 
than  5. 

68.  To  find  the  Logarithm  of  a  Given  Number.  —  When 
the  given  number  has  not  more  than  five  digits,  we  have 
merely  to  take  the  mantissa  immediately  from  the  table, 
and  prefix  the  characteristic  by  the  rule  (Art.  64). 

Thus,  suppose  we  require  the  logarithm  of  62541.  The 
table  gives  .7961648  as  the  mantissa,  and  the  characteristic 
is  4,  by  the  rule ;  therefore 

log    62540.  =  4.7961648. 
Similarly,  log  .006^81  =  3.7980288     .     .     (Art.  64) 


100  PLANE  TRIGONOMETRY. 

Suppose,  however,  that  the  given  number  has  more  than 
five  digits.  For  example : 

Suppose  we  require  to  find  log  62761.6. 
We  find  from  the  table 

log  62761  =  4. 7976899 

log  62762  =  4.7976968 

and  diff.  for  1  =  0.0000069 

Thus  for  an  increase  of  1  in  the  number  there  is  an  in- 
crease of  .0000069  in  the  logarithm. 

Hence,  assuming  that  the  increase  of  the  logarithm  is 
proportional  to  the  increase  of  the  number,  then  an  increase 
in  the  number  of  .6  will  correspond  to  an  increase  in  the 
logarithm  of  .6  x  .0000069  =  .0000041,  to  the  nearest  sev- 
enth decimal  place. 

Hence,  log  62761  =  4.7976899 

diff.  for  .6  =  41 


.-.  log  62761.6  =  4.7976940 

This  explains  the  use  of  the  column  of  proportional  parts 
on  the  extreme  right  of  the  page.  It  will  be  seen  that  the 
difference  between  the  logarithms  of  two  consecutive  num- 
bers is  not  always  the  same;  for  instance,  those  in  the 
upper  part  of  the  page  before  us  differ  by  .0000070,  while 
those  in  the  middle  and  the  lower  parts  differ  by  .0000069 
and  .0000068.  Under  the  column  with  the  heading  69  we 
see  the  difference  41  corresponding  to  the  figure  6,  which 
implies  that  when  the  difference  between  the  logarithms  of 
two  consecutive  members  is  .0000069,  the  increase  in  the 
logarithm  corresponding  to  an  increase  of  .6  in  the  number 
is  .0000041 ;  for  .06  it  is  evidently  .0000004,  and  so  on. 

NOTE.  —  We  assume  in  this  method  that  the  increase  in  a  logarithm  is  propor- 
tional to  the  increase  in  the  number.  Although  this  is  not  strictly  true,  yet  it  is  in 
most  rasps  fiiifticiently  exact  for  practical  purposes. 

Had  we  taken  a  whole  number  or  a  decimal,  the  process  would  have  been  the 
same. 


TABLE  OF 


1,01 


N. 

O 

1 

2 

3 

4 

5 

6 

9217 

7 

8  1  9 

F 

1 
2 
3 
4 

5 
6 
7 
8 
9 

1 
2 
3 
4 

5 
6 

7 
8 

9 

1 
2 
3 
4 
5 
6 
7 
8 
9 

'.P. 

6250 

51 
52 
53 
54 
55 
56 
57 
58 
59 
6260 
61 
62 
63 
64 
65 
66 
67 
68 
69 
6270 
71 
72 
73 
74 
75 
76 
77 
78 
79 

6280 

81 
82 
83 
84 
85 
86 
87 
88 
89 
6290 
91 
92 
93 
94 
95 
96 
97 
98 
99 
6300 

795  8800 

8870 

8939 

9009 

9078 

9148 

9,^£v 

53A6 

0051 
0745 
1440 
2134 
2829 
3523 
4217 
4911 
5605 

9126 

0120 
0815 
1509 
2204 
2898 
3592 
4286 
4980 
5674 
6368 

70 
7.0 
14.0 
21.0 
28.0 
35.0 
42.0 
49.0 
56.0 
63.0 

69 
6.9 
13.8 
20.7 
27.6 
34.5 
41.4 
48.3 
55.2 

62.1 

68 
6.8 
13.6 
20.4 
27.2 
34.0 
40.8 
47.6 
54.4 
61.2 

9495 
796  0190 
0884 
1579 
2273 
2967 
3662 
4356 
5050 

9564 
0259 
0954 
1648 
2343 
3037 
3731 
4425 
5119 

9634 
0329 
1023 
1718 
2412 
3106 
3800 
4494 
5188 

9703 
0398 
1093 
1787 
2481 
3176 
3870 
4564 
5258 

9773 
0468 
1162 
1857 
2551 
3245 
3939 
4633 
5327 

9842 
0537 
1232 
1926 
2620 
3314 
4009 
4703 
5396 

9912 
0606 
1301 
1995 
2690 
3384 
4078 
4772 
5466 

9981 
0676 
1370 
2065 
2759 
3453 
4147 
4841 
5535 

7965743 

5813 

5882 

5951 
6645 
7339 
8032 
8725 
9419 
0112 
0805 
1498 
2191 

6021 

6090 

6160 

6229 

6298 
6992 
7685 
8379 
9072 
9765 
0458 
1151 
1844 
2537 

6437 
7131 
7824 
8517 
9211 
9904 
797  059  7 
1290 
1983 

6506 
7200 
7893 
8587 
9280 
9973 
0666 
1359 
2052 

6576 
7269 
7963 
8656 
9349 
0043 
0736 
1428 
2121 

6714 
7408 
8101 
8795 
9488 
0181 
0874 
1567 
2260 

6784 
7477 
8171 
8864 
9557 
0250 
0943 
1636 
2329 

6853 
7547 
8240 
8933 
9627 
0320 
1013 
1706 
2398 

6923 
7616 
8309 
9003 
9696 
0389 
1082 
1775 
2468 
3160 
3853 
4545 
5237 
5930 
6622 
7314 
8006 
8697 
9389 

7061 
7755 
8448 
9141 
9835 
0528 
1221 
1913 
2606 

797  2675 

27452814 

2883 

2952 

3022 

3091 

3229 

3299 

3368 
4060 
4753 
5445 
6137 
6829 
7521 
8213 
8905 

797  9596 

3437 
4130 
4822 
5514 
6207 
6899 
7590 
8282 
8974 

9666 

3507 
4199 
4891 
5584 
6276 
6968 
7660 
8351 
9043 

9735 
0426 
1118 
1809 
2500 
3191 
3882 
4573 
5263 
5954 

3576 
4268 
4961 
5653 
6345 
7037 
7729 
8421 
9112 

9804 

3645 
4337 
5030 
5722 
6414 
7106 
7798 
8490 
9181 

9873 

3714 
4407 
5099 
5791 
6483 
7175 
7867 
8559 
9251 

3784 
4476 
5168 
5860 
6553 
7245 
7936 
8628 
9320 

3922 
4614 
5307 
5999 
6691 
7383 
8075 
8766 
9458 

3991 
4684 
5376 
6068 
6760 
7452 
8144 
8836 
9527 

9942 

0011 

0080 

0150 

6219 
0910 
1601 
2293 
2984 
3675 
4366 
5056 
5747 
6437 

798  0288 
0979 
1671 
2362 
3053 
3744 
4435 
5125 
5816 

0357 
1048 
1740 
2431 
3122 
3813 
4504 
5194 
5885 

0495 
1187 
1878 
2569 
3260 
3951 
4642 
5333 
6023 

0565 
1256 
1947 
2638 
3329 
4020 
4711 
5402 
6092 

0634 
1325 
2016 
2707 
3398 
4089 
4780 
5471 
6161 

0703 
1394 
2085 
2776 
3467 
4158 
4849 
5540 
6230 

0772 
1463 
2154 
2846 
3536 
4227 
4918 
5609 
6299 

0841 
1532 
2224 
2915 
3606 
4296 
4987 
5678 
6368 

798  6506 

6575 

6645 

6714 

6783 

6852 

6921 

6990 
7680 
8370 
9060 
9750 
0440 
1130 
1820 
2509 
3199 

7059 
7749 
8439 
9129 
9819 
0509 
1199 
1889 
2578 
3268 

7128 

7197 
7887 
8577 
9267 
9957 
799  0647 
1337 
2027 
2716 
799  3405 

7266 
7956 
8646 
9336 
0026 
0716 
1406 
2096 
2785 

7335 
8025 
8715 
9405 
0095 
0785 
1475 
2164 
2854 

7404 
8094 
8784 
9474 
0164 
0854 
1544 
2233 
2923 

7473 
8163 
8853 
9543 
0233 
0923 
1613 
2302 
2992 

7542 
8232 
8922 
9612 
0302 
0992 
1682 
2371 
3061 

7611 
8301 
8991 
9681 
0371 
1061 
1751 
2440 
3130 
3819 

7818 
8508 
9198 
9888 
0578 
1268 
1958 
2647 
3337 

3474 

3543 

3612 

3681 

3750 

3888 

3957 

4026 

N. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

P.P. 

102  PLANE  TRIGONOMETRY. 

Thus,' suppose'  we  require  to  find  log  627616  and  log  .627616.  The  mantissa  is 
exactly  the  same  as  before  (Art.  66),  and  the  only  difference  to  be  made  in  the  final 
result  is  to  change  the  characteristic  according  to  rule  (Art.  64). 

Thus  log  627616  =  5.7976942, 

and  log  .627616  =  1.7976942. 

69.  To  find  the  Number  corresponding  to  a  Given  Loga- 
rithm. —  If  the  decimal  part  of  the  logarithm  is  found  ex- 
actly in  the  table,  we  can  take  out  the  corresponding 
number,  and  put  the  decimal  point  in  the  number,  in  the 
place  indicated  by  the  characteristic. 

Thus  if  we  have  to  find  the  number  whose  logarithm  is 
2.7982915,  we  look  in  the  table  for  the  mantissa  .7982915, 
and  we  find  it  set  down  opposite  the  number  62848 :  and 
as  the  characteristic  is  2,  there  must  be  one  cipher  before 
the  first  significant  figure  (Art.  64). 

Hence  2.7982915  is  the  logarithm  of  .062848. 

Next,  suppose  that  the  decimal  part  of  the  logarithm'  is 
not  found  exactly  in  the  table.  For  example,  suppose  we 
have  to  find  the  number  whose  logarithm  is  2.7974453. 

We  find  from  the  table 

log  62726  =  4.7974476 
log  62725  =  4.7974407 
diff .  for  1  =  .0000069 

Thus  for  a  difference  of  1  in  the  numbers  there  is  a 
difference  of  .0000069  in  the  logarithms.  The  excess  of 
the  given  mantissa  above  .7974407  is  (.7974453  -  .7974407) 
or  .0000046. 

Hence,  assuming  that  the  increase  of  the  number  is 
proportional  to  the  increase  of  the  logarithm,  we  have 

.0000069  :  .0000046  :  :  1  :  number  to  be  added  to  627.25. 

.0000046      46       ™  69) 46.0 (.666 
.-.  number  to  be  added  =  -          —  =  —  =  .667 

.0000069      69  41 4 

.-.  log  62725.667  =  4.7974453,  ^60 

and       .-.  log 627.25667  =  2.7974453;  414 

therefore  number  required  is  627.25667.  460 


ARITHMETIC  COMPLEMENT.  103 

We  might  have  saved  the  labor  of  dividing  46  by  69,  by 
using  the  table  of  proportional  parts  as  follows : 

given  mantissa  =  .7974453 
mantissa  of  62725  =  .7974407 
diff.  of  mantissse  =  46 

proportional  part  for  .6      =  41.4 

4.6 
"  "      "   .06    =  4.14 


.46 
«  «      «   .006=  .414 

and  so  on. 
.-.  number  =  627.256666-... 

69  a.  Arithmetic  Complement,  —  By  the  arithmetic  com- 
plement of  the  logarithm  of  a  number,  or,  briefly,  the 
cologarithm  of  the  number,  is  meant  the  remainder  found 
by  subtracting  the  logarithm  from  10.  To  subtract  one 
logarithm  from  another  is  the  same  as  to  add  the  Co- 
logarithm  and  then  subtract  10  from  the  result. 

Thus,  a  -  b  =  a  +  (10  -  b)  -  10, 

where  a  and  b  are  logarithms,  and  10  —  6  is  the  arithmetic 
complement  of  b. 

When  one  logarithm  is  to  be  subtracted  from  the  sum  of 
several  others,  it  is  more  convenient  to  add  its  cologarithm 
to  the  sum,  and  reject  10.  The  advantage  of  using  the 
cologarithm  is  that  it  enables  us  to  exhibit  the  work  in  a 
more  compact  form. 

The  cologarithm  is  easily  taken  from  the  table  mentally 
by  subtracting  the  last  significant  figure  on  the  right  from 
10,  and  all  the  others  from  9. 


104 


PLANE  TRIGONOMETRY. 


1.  Given 

find 

2.  Given 

find 

3.  Given 

find 

4.  Given 

find 

5.  Given 


EXAMPLES. 

log  52502  =4.7201758, 
log  52503  =4.7201841; 
log  52502.5. 

log  3.0042  =  0.4777288, 
log  3.0043  =  0.4777433; 
log  300.425. 

log  7.6543  =  0.8839055, 
log  7.6544  =  0.8839112; 
log  7.65432. 

log  6.4371  =  0.8086903, 
log  6.4372  =  0.8086970; 
log  6437125. 


Ans.  4.7201799. 


2.4777360. 


.8839066. 


log  12954  =4.1124039, 
log  12955  =4.1124374; 
find  the  number  whose  logarithm  is  4.1124307. 

6.  Given  log  60195  =  4.7795532, 

log  60196  =4.7795604; 
find  the  number  whose  logarithm  is  2.7795561. 

7.  Given  log  3.7040  =  .5686710, 

log  3.7041  =  .5686827; 
find  the  number  whose  logarithm  is  .5686760. 

8.  Given  log  2.4928  =  .3966874, 

log  2.4929  =  .3967049; 
find  the  number  whose  logarithm  is  6.3966938. 


6.8086920. 


12954.8. 


601.95403. 


3.70404. 


2492837. 


NATURAL   TRIGONOMETRIC  FUNCTIONS.        105 

9.   Given  log  32642  =  4.5137768, 

log  32643  =4.5137901; 
find  log  32642.5.  Arts.  4.5137835. 

10.  Find  the  logarithm  of  62654326.  7.7969510. 
Use  specimen  page. 

11.  Find  the  number  whose  logarithm  is  4.7989672. 

Ana.  62945.876. 

70.  Use  of  Trigonometric  Tables. —Trigonometric  Tables 
are  of  two  kinds,  — Tables  of  Natural  Trigonometric  Functions 
and  Tables  of  Logarithmic  Trigonometric  Functions.  As  the 
greater  part  of  the  computations  of  Trigonometry  is  carried 
on  by  logarithms,  the  latter  tables  are  by  far  the  most  use- 
ful. 

We  have  explained  in  Art.  27  how  to  find  the  actual 
numerical  values  of  certain  trigonometric  functions,  exactly 
or  approximately. 

Thus,      sin  30°  =  1;  that  is,  .5  exactly. 
2i 

Also,       tan  60°  =  V3 ;  that  is,  1.73205  approximately. 

A  table  of  natural  trigonometric  functions  gives  their 
approximate  numerical  values  for  angles  at  regular  intervals 
in  the  first  quadrant.  In  some  tables  the  angles  succeed 
each  other  at  intervals  of  1",  in  others,  at  intervals  of  10", 
but  in  ordinary  tables  at  intervals  of  1' :  and  the  values  of 
the  functions  are  given  correct  io  five,  six,  and  seven  places. 
The  functions  of  intermediate  angles  can  be  found  by  the 
principle  of  proportional  parts  as  applied  in  the  table  of 
logarithms  of  numbers  (Arts.  68  and  69). 

It  is  sufficient  to  have  tables  which  give  the  functions 
of  angles  only  in  the  first  quadrant,  since  the  functions  of 
all  angles  of  whatever  size  can  be  reduced  to  functions 
of  angles  less  than  90°  (Art.  35). 


106  PLANE   TRIGONOMETRY. 

71.  Use  of  Tables  of  Natural  Trigonometric  Functions.  — 

These  tables,  which  consist  of  the  actual  numerical  values 
of  the  trigonometric  functions,  are  commonly  called  tables 
of  natural  sines,  cosines,  etc.,  so  as  to  distinguish  them  from 
the  tables  of  the  logarithms  of  the  sines,  cosines,  etc. 

We  shall  now  explain,  first,  how  to  determine  the  value 
of  a  function  that  lies  between  the  functions  of  two  con- 
secutive angles  given  in  the  tables ;  and  secondly,  how  to 
determine  the  angle  to  which  a  given  ratio  corresponds. 

72.  To  find  the  Sine  of  a  Given  Angle. 

Find  the  sine  of  25°  14'  20",  having  given  from  the  table 

sin  25°  15' =  .4265687 
sin  25°  14'  =  .4263056 


diff.  for  1'  ==  .0002631 

Let  d  =  diff.  for  20" ;  and  assuming  that  an  increase  in 
the  angle  is  proportional  to  an  increase  in  the  sine,  we  have 

60  :  20  :  :  .0002631  :  d. 

^  =  20  x. 0002631 

60 

.-.  sin  25°  14'  20"  =  .4263056  +  .0000877 
=  .4263933. 

NOTE.  —  We  assumed  here  that  an  increase  in  the  angle  is  proportional  to  the 
increase  in  the  corresponding  sine,  which  is  sufficiently  exact  for  practical  purposes, 
with  certain  exceptions. 

73.   To  find  the  Cosine  of  a  Given  Angle. 

Find  the  cosine  of  44°  35'  25",  having  given  from  the  table 

cos  44°  35'  =  .7122303 
cos  44°  36'  =  .7120260 


diff.  for  1'  =  .0002043 

observing  that  the  cosine  decreases  as  the  angle  increases 
from  0°  to  90°. 


EXAMPLES. 


107 


Let  d  =  decrease  of  cosine  for  25"  ;  then 
60  :  25  :  :  .0002043  :  d. 

f)K 

...  d  =  —x  .0002043  =  .0000851. 

.*.  cos  44°  35'  25"  =  .7122303  -  .0000851 
=  7121452. 

Similarly,  we  may  find  the  values  of  the  other  trigono- 
metric functions,  remembering  that,  in  the  first  quadrant, 
the  tangent  and  secant  increase  and  the  cotangent  and 
cosecant  decrease,  as  the  angle  increases. 


1.  Given 

find 

2.  Given 

find 

3.  Given 

find 

4.  Given 

find 

5.  Given 

find 


EXAMPLES. 

sin  44°  35'=  .7019459, 
sin  44°  36'=  .7021531; 
sin  44°  35'  25". 

sin  42°  15' =.6723668, 
sin  42°  16' =.6725821; 
sin  42°  15'  16". 

sin  43°  23'=  .6868761, 
sin  43°  22'=  .6866647; 
sin  43°  22' 50". 

sin  31°  6'  =.5165333, 
sin  31°  7'  =.5167824; 
sin  31°  6'  25". 

cos  74°  45'=  .4265687, 
cos  74°  46' =.4263056; 
cos  74°  45'  40". 


Ans.  .7020322. 


.6724242. 


.6868408. 


.5166371. 


.4263933. 


108  PLANE   TRIGONOMETRY. 

6.  Given  cos  41°  13'=  .7522233, 

cos  41°  14'=  .7520316; 
find  cos  41°  13' 26".  Ans.  .7521403. 

7.  Given  cos  47°  38'=  .6738727, 

cos  47°  39'=  .6736577; 
find  cos  47°  38'  30".  .6737652. 

74.  To  find  the  Angle  whose  Sine  is  Given. 

Find  the  angle  whose  sine  is  .5082784,  having  given  from 
the  table  gin  3Qo  33,  =  5082901 

sin  30°  32' =  .5080396 
diff.  for  1'  =  . 0002505 

given  sine  =  .5082784 
sin  30°  32'  =  .5080396 

diff.  =  .0002388 

Let  d  =  diff.  between  30°  32'  and  required  angle;   then 
.0002505  :  .0002388  : :  60  :  d. 

d  =  2388  x  60  =  6552 
2505        ~  167 

=  57.2  nearly. 
.-.  required  angle  =  30°  32'  57". 2. 

75.  To  find  the  Angle  whose  Cosine  is  Given. 

Find  the  angle  whose  cosine  is  .4043281,  having  given 
from  the  table          cos  66°    9' =  .4043436 
cos  66°  10'  =  .4040775 
diff.  for  1'  =  .0002661 

cos  66°  9'  =  .4043436 
given  cosine  =  .4043281 

diff.  =  .0000155 


EXA3IP 


109 


Let   d  =  diff.  between  66°  9'  and  required  angle;   tken 
.0002661:  .0000155  :  :  60  :  d. 


Required  angle  is  greater  than  66°  91  because  its  cosine  is 
less  than  cos  66°  9'. 

.-.  required  angle  =66°  9'  3".5. 


EXAMPLES 

1.  Given  sin  44°  12' =  .6971651, 

sin  44°  11'=.  6969565; 
find  the  angle  whose  sine  is  .6970886.  Ans.  44°  11'  38". 

2.  Givej  sin  48°  47'  =  . 7522233, 

sin  48°  46' =  .7520316; 
find  the  angle  whose  sine  is  .752140.  48°  46'  34". 

3.  Given  sin  24°  11' =  .4096577. 

sin  24°  12'  =  .4099230; 
find  the  angle  whose  sine  is  .4097559.  24°  11'  22".2. 

4.  Given  cos  32°  3T  =  .8432351, 

cos  32°  32*  =  .£430787; 
find  the  angle  whose  cosine  is  .8432.  32°  31*  13".5. 

5.  Given  cos  44°  11' =  .7171134, 

cos  44°  12*  =  . 7169106; 
find  the  angle  whose  cosine  is  .7169848.  44°  11'  38". 

6.  Given  cos  70°  32'  =  .3332584, 

€0870-31'  =  . 3335326; 
find  the  angle  whose  cosine  is  .3333333.  79°  31'  43".6. 


110  PLANE  TRIGONOMETRY. 

76.  Use  of  Tables  of  Logarithmic  Trigonometric  Func- 
tions. —  Since  the  sines,  cosines,  tangents,  etc.,  of  angles 
are  numbers,  we  may  use  the  logarithms  of  these  numbers 
in  numerical  calculations  in  which  trigonometric  functions 
are  involved;  and  these  logarithms  are  in  practice  much 
more  useful  than  the  numbers  themselves,  as  with  their 
assistance  we  are  able  to  abbreviate  greatly  our  calcula- 
tions ;  this  is  especially  the  case,  as  we  shall  see  hereafter, 
in  the  solution  of  triangles.  In  order  to  avoid  the  trouble 
of  referring  twice  to  tables — first  to  the  table  of  natural 
functions  for  the  value  of  the  function,  and  then  to  a  table 
of  logarithms  for  the  logarithm  of  that  function  —  the  log- 
arithms of  the  trigonometric  functions  have  been  calculated 
and  arranged  in  tables,  forming  tables  of  the  logarithms  of 
the  sines,  logarithms  of  the  cosines,  etc. ;  these  tables  are 
called  tables  of  logarithmic  sines,  logarithmic  cosines,  etc. 

Since  the  sines  and  cosines  of  all  angles  and  the  tangents 
of  angles  less  than  45°  are  less  than  unity,  the  logarithms  of 
these  functions  are  negative.  To  avoid  the  inconvenience  of 
using  negative  characteristics,  10  is  added  to  the  logarithms 
of  all  the  functions  before  they  are  entered  in  the  table. 
The  logarithms  so  increased  are  called  the  tabular  logarithms 
of  the  sine,  cosine,  etc.  Thus,  the  tabular  logarithmic  sine 
of  30°  is 

10  +  log  sin  30°  =  10  +  logi  =  10  -  log  2  =  9.6989700. 

In  calculations  we  have  to  remember  and  allow  for  this 
increase  of  the  true  logarithms.  When  the  value  of  any 
one  of  the  tabular  logarithms  is  given,  we  must  take  away 
10  from  it  to  obtain  the  true  value  of  the  logarithm. 

Thus  in  the  tables  we  find 

log  sin  31°  15'  =  9.7149776. 

Therefore  the  true  value  of  the  logarithm  of  the  sine  of 
31°  15'  is  9.7149776  -  10  =  1.7149776. 

Similarly  with  the  logarithms  of  other  functions. 


TABLES  OF  LOGARITHMIC  FUNCTIONS.         Ill 

NOTE.  —  English  authors  usually  denote  these  tabular  logarithms  by  the  letter  L 
Thus,  Lsin  A  denotes  the  tabular  logarithm  of  the  sine  of  A. 

French  authors  use  the  logarithms  of  the  tables  diminished  by  10.    Thus, 

log  sin  A  =  1.8598213,  instead  of  9.8598213. 

The  Tables  contain  the  tabular  logs  of  the  functions  of  all 
angles  in  the  first  quadrant  at  intervals  of  1' ;  and  from 
these  the  logarithmic  functions  of  all  other  angles  can  be 
found.* 

Since  every  angle  between  45°  and  90°  is  the  complement 
of  another  angle  between  45°  and  0°,  every  sine,  tangent, 
etc.,  of  an  angle  less  than  45°  is  the  cosine,  cotangent,  etc., 
of  another  angle  greater  than  45°  (Art.  16).  Hence  the 
degrees  at  the  top  of  the  tables  are  generally  marked  from 
0°  to  45°,  and  those  at  the  bottom  from  45°  to  90°,  while 
the  minutes  are  marked  both  in  the  first  column  at  the  left, 
and  in  the  last  column  at  the  right.  Every  number  there- 
fore in  eacji  column,  except  those  marked  diff.,  stands  for 
two  functions — the  one  named  at  the  top  of  the  column, 
and  the  complemental  function  named  at  the  bottom  of  the 
column.  In  looking  for  a  function  of  an  angle,  if  it  be  less 
than  45°,  the  degrees  are  found  at  the  top,  and  the  minutes 
at  the  left-hand  side.  If  greater  than  45°,  the  degrees  are 
found  at  the  foot,  and  the  minutes  at  the  right-hand  side. 

On  page  113  is  a  specimen  page  of  Mathematical  Tables. 
It  gives  the  tabular  logarithmic  functions  of  all  angles  between 
38°  and  39°,  and  also  of  those  between  51°  and  52°,  both 
inclusive,  at  intervals  of  1'.  The  names  of  the  functions 
for  38°  are  printed  at  the  top  of  the  page,  and  those  for  51° 
at  the  foot.  The  column  of  minutes  for  38°  is  on  the  left, 
that  for  51°  is  on  the  right. 

Thus  we  find 

log  sin  38°  29'  =9.7939907. 
log  cos  38°  45'  =  9.8920303. 
log  tan  51°  18'  =  10.0962856. 

*  Many  tables  are  calculated  for  angles  at  intervals  of  10". 


112  PLANE   TRIGONOMETRY. 

77.  To  find  the  Logarithmic  Sine  of  a  Given  Angle. 

Find  log  sin  38°  52'  46". 
We  have  from  page  113 

log  sin  38°  53'  =  9.7977775 

log  sin  38°  52'  =  9.7976208 

cliff,  for  1'  =    .0001567 

Let  d  —  diff.  for  46",  and  assuming  that  the  chancre  in 
the  log  sine  is  proportional  to  the  change  in  the  angle,  we 

have 

60  :  46  :  :  .0001567  :  d. 

d=s  46  x.  0001567  ^0001201 

60 

.-.     log  sin  38°  52'  46"=  9.7976208  +  .0001201 
=  9.7977409. 

78.  To  find  the  Logarithmic  Cosine  of  a  Given  Angle. 

Find  log  cos  83°  27'  23",  having  given  from  the  table 

log  cos  83°  27'=  9.0571723 
log  cos  83°  28'=  9.0560706 

diff.  for  1'=    .0011017 
Let  d  =  decrease  of  log  cosine  for  23"  ;  then 

60  :  23  :  :  .0011017  :  d. 
23x1017 


.-.     log  cos  83°  27'  23"=  9.0571723  -  .0004223 
=  9.0567500. 

EXAMPLES. 

1.    Giren      log  sin    6°  33'=  9.0571723, 
log  sin    6°  32'  =9.0560706; 
find  log  sin    6°  32'  37".  Ana.  9.05675. 


38  Deg. 


TABLE  OF  LOGARITHMS. 


113 


; 

Sine. 

Diff. 

Tang. 

Diff. 

Cotang. 

Diff. 

Cosine. 

f 

o 

I 

2 

3  i 
4 
5 

9.7893420 
9.7895036 
9.7896652 
9.7898266 
9.7899880 
9.7901493 

1616 
1616 
1614 
1614 
1613 

9.8928098 
9.8930702 
9-8933306 
9.8935909 
9.8938511 
9.8941114 

2604 
2604 
2603 
2602 
2603 

10.1071902 
10.1069298 
10.1066694 
10.1064091 
10.1061489 
10.1058886 

987 
988 
988 
989 
990 

QQO 

9.8965321 
9.8964334 
9.8963346 
9.8962358 
9.8961369 
9.8960379 

60 
59 

58 

55 

6 

7 

8 

9 
10 

9.7903104 
9.7904715 
9.7906325 
9.7907933 
9.7909541 

1611 
1610 
1608 
1608 
1607 

9.8943715 
9.8946317 
9.8948918 
9.8951519 
9.8954119 

2602 
2601 
2601 
2600 
2600 

10.1056285 
10.1053683 
10.1051082 
10.1048481 
10.1045881 

991 
992 
992 
992 

QQO 

9-8959389 
9.8958398 
9.8957406 
9.8956414 
9.8955422 

54 
53 
52 

So 

ii 

12 
13 
14 
15 

9.7911148 
9.7912754 
97914359 
9-7915963 
9.7917566 

1606 
1605 
1604 
1603 

9.8956719 

9-89593J9 
9.8961918 
9.8964517 
9.8967116 

2600 
2599 
2599 
2599 

2CQ8 

10.1043281 
10.1040681 
10.1038082 
10.1035483 
10.1032884 

994 
995 
995 
995 

QQ7 

9.8954429 

9-8953435 
9.8952440 
9.8951445 
9.8950450 

49 
48 

45 

16 

17 
18 

19 

20 

9.7919168 
9.7920769 
9.7922369 
9.7923968 
9.7925566 

1601 
1600 
1599 
1598 

9.8969714 
9.8972312 
9.8974910 
9.8977507 
9.8980104 

2598 
2598 
2597 
2597 

10.1030286 
10.1027688 
10.1025090 
10.1022493 
10.1019896 

996 
998 
998 
998 
IOOO 

9.8949453 
9.8948457 
9.8947459 
9.8946461 
9.8945463 

44 
43 
42 

40 

21 
22 
23 
24 
25 

9.7927163 
9.7928760 
97930355 
97931949 
9.7933543 

1597 
1595 
1594 
1594 

9.8982700 
9.8985296 
9.8987892 
9.8990487 
9.8993082 

2596 
2596 
2595 
2595 

2CQC 

10.1017300 
10.1014704 
10.1012108 
10.1009513 
10.1006918 

999 

IOOI 
IOOI 
IOOI 
1003 

9-8944463 
9.8943464 
9.8942463 
9.8941462 
9.8940461 

P 
P 

35 

26 

27 

28 
29 

3° 

9.7935I35 
9.7936727 

9-79383I7 
9.7939907 
9.7941496 

1589 

9.8995677 

9.9003459 
9.9006052 

2594 
2594 
2594 
2593 

10.1004323 
10.1001729 
10.0999135 
10.0996541 
10.0993948 

1002 

1004 
1004 
1004 

9.8939458 
9.8938456 
9.8937452 
9.8936448 
9.8935444 

34 
33 
32 
31 
30 

32 
33 
34 
35 

9.7943083 
9.7944670 
9-7946256 
9.7947841 
9.7949425 

1587 
1586 
1585 
1584 

9.9008645 
9.9011237 
9.9013830 
9.9016422 

2592 
2593 
2592 

2591 

2CQI 

10.0991355 

10.0988763 
10.0986170 
10.0983578 
10.0980987 

1006 
1007 
1007 
1007 
1008 

9.8934439 

9.8933433 
9-8932426 
9.8931419 
9.8930412 

27 
26 
25 

36 
37 
38 
39 
40 

9.7951008 
9.7952590 
9.7954171 
9.7955751 
9-7957330 

1582 

9.9021604 
9.9024195 

9.9029376 
9.9031966 

2591 
2591 
2590 
2590 

10.0978396 
10.0975805 
10.0973214 
10.0970624 
10.0968034 

1009 

1010 
1010 
1010 
IOII 

9.8929404 
9.8928395 
9.8927385 
9.8926375 
9.8925365 

24 
23 

22 
21 
2O 

4i 

42 

43 
44 

45 

9.7958909 
9.7960486 
9.7962062 
9-7963638 
9.7965212 

1 

9.9034555 
9.9037144 

9.9039733 
9.9042321 

2589 
2589 
2588 
2589 
2^87 

10.0965445 
10.0962856 
10.0960267 
10.0957679 
10.0955090 

1012 
1013 
1013 
1013 

IOI4 

9.8924354 
9.8923342 
9.8922329 
9.8921316 
9.8920303 

19 

18 

17 

16 
15 

46 

47 
48 

49 
50 

9.7966786 
9-7968359 
9.7969930 
9.7971501 
9.7973071 

s? 

1571 
1570 

1560 

9.9047497 
9.9050085 
9.9052672 
9.9055259 
9.9057845 

2588 
2587 

2c;86 

10.0952503 
10.0949915 
10.0947328 
10.0944741 
10.0942155 

IOI5 

1016 
1016 
1016 

1018 

9.8919289 
9.8918274 
9.8917258 
9.8916242 
9.8915226 

14 
13 

12 
II 
10 

52 
53 

54 
55 

9.7974640 

9.7976208 
9.7977775 

9.7979341 
9.7980906 

1568 

1567 
1566 

'565 

9.9060431 
9.9063017 
9.9065603 
9.9068188 
9.9070773 

2586 
2586 
2585 
2585 
2^84 

10.0939569 
10.0936983 
10.0934397 
10.0931812 
10.0929227 

1017 
1019 
1019 

IO2O 

9.8914208 
9.8913191 
9.8912172 
9.8911153 
9.8910133 

9 
8 

7 
6 

5 

56 
1 

9.7982470 
9.7984034 
97985596 
9.7987158 
9.7988718 

1564 

1562 
1562 
1560 

9-9073357 
9.907594I 
9.9078525 
9.9081109 
9.9083692 

2584 

2584 
2584 

2583 

10.0926643 
10.0924059 
10.0921475 
10.0918891 
10.0916308 

IO2I 
IO2I 
IO22 
1023 

9.8909113 
9.8908092 
9.8907071 
9.8906049 
9.8905026 

4 
3 

2 

I 
O 

t 

Cosine. 

Diff. 

Cotang, 

Diff. 

Tang. 

Diff. 

Sine, 

t 

51  Deer. 


114 


PLANE   TRIGONOMETRY. 


2.  Given      log  sin  55°  33'=  9.9162539, 

log  sin  55°  34'=  9.9163406 ; 
find  log  sin  55°  33' 54".  Ans.  9.9163319. 

3.  Given      log  cos  37°  28'=  9.8996604, 

log  cos  37°  29'=  9.8995636 ; 
find  log  cos  37°  28'  36".  9.8996023. 

4.  Given      log  cos  44°  35'  20"=  9.8525789, 

log  cos  44°  35'  30"=  9.8525582 ; 

find  log  cos  44°  35'  25".7.  9.8525671. 

See  foot-note  of  Art.  76. 

5.  Given      log  cos  55°  11'=  9.7565999, 

log  cos  55°  12'=  9.7564182 ; 


find 


log  cos  55°  11' 12". 


6.    Given      log  tan  27°  13'=  9.7112148, 
log  tan  27°  14'=  9.7115254 ; 


find 


log  tan  27°  13' 45". 


9.7565636. 


9.7114477. 


79.   To  find  the  Angle  whose  Logarithmic  Sine  is  Given. 

Find  the  angle  whose  log  sine  is  8.8785940,  having  given 

from  the  table 

log  sin  4°  21 '=8.8799493 

log  sin  4°  20'  =  8.8782854 
diff.  for  1'=    .0016639 

given  log  sine  =  8.8785940 

log  sin  4°  20'=  8.8782854 

diff.  =    .0003086 

Let  d  =  diff.  between  4°  20'  and  required  angle  ;  then 
.0016639  :  .0003086  :  :  60  :  d. 
3086  x  60 


cl  = 


=  24,  nearly. 


16639 
.-.  required  angle  =  4°  20'  24". 


EXAMPLES.  115 

80.  To  find  the  Angle  whose  Logarithmic  Cosine  is  Given. 

Find  the  angle  whose  log  cosine  is  9.8934342. 
^Ve  have  from  page  113 

log  cos  38°  31'=  9.8934439 
log  cos  38°  32'=  9.8933433 

diff.  for  1'=    .0001006 

log  cos  38°  31'=  9.8934439 
given  log  cosine  =  9.8934342 

diff.  =    .0000097 
Let  d  =  diff.  between  38°  31'  and  required  angle  ;  then 

.0001006  :  .0000097  :  :  60  :  d. 
.  ,0000097  x 


.. 

.0001006  1006 

.-.  required  angle  =  38°  31'  5".8. 

XOTE.  —  In  using  both  the  tables  of  the  natural  sines,  cosines,  etc.,  and  the  tables 
of  the  logarithmic  sine.-*,  cosines,  etc.,  the  student  will  remember  that,  in  the  first 
quadrant,  as  the  angle  increases,  the  sine,  tangent,  and  secant  increase,  but  the 
cosine,  cotangent,  and  cosecant  decrease. 

EXAMPLES. 

1.  Given         log  sin  14°  24'=  9.3956581, 

log  sin  14°  25'  -9.3961499; 
find  the  angle  whose  log  sine  is  9.3959449.    Ans.  14°  24'  35". 

2.  Given        log  sin  71°  40'=  9.9773772, 

log  sin  71°  41'  =9.9774191; 
fmd  the  angle  whose  log  sine  is  9.9773897.  71°  40'  18". 

3.  Given        log  cos  28°  17'=  9.9447862, 

log  cos  28°  16'  =9.9448541; 
find  the  angle  whose  log  cosine  is  9.9448230.      28°  16'  27".5. 


116  PLANE   TRIGONOMETRY. 

4.  Given         log  cos  80°  53'=  9.1998793, 

log  cos  80°  52'  50"=  9.2000105 ; 
find  the  angle  whose  log  cosine  is  9.2000000. 

Ans.  80°  52' 51". 

5.  Given        log  tan  35°  4'=  9.8463018, 

log  tan  35°  5'=  9.8465705; 
find  the  angle  whose  log  tangent  is  9.8464028.        35°  4'  23". 

6.  Given        log  sin  44°  35'  30"=  9.8463678, 

log  sin  44°  35'  20"=  9.8463464; 
find  the  angle  whose  log  sine  is  9.8463586.          44°  35'  25". 7. 

7.  Find   the   angle   by   page    113  whose   log  tangent   is 
10.1018542.  Ans.  51°  39' 28". 7. 

81.  Angles  near  the  Limits  of  the  Quadrant,  —  It  was 

assumed  in  Arts.  72-80  that,  in  general,  the  differences  of 
the  trigonometric  functions,  both  natural  and  logarithmic, 
are  approximately  proportional  to  the  differences  of  their 
corresponding  angles,  with  certain  exceptions.  The  excep- 
tional cases  are  as  follows : 

(1)  Natural  functions.  —  For  the  sine  the  differences  are 
insensible  for  angles  near  90°;  for  the  cosine  they  are  in- 
sensible for  angles  near  0°.     For  the  tangent  the  differences 
are  irregular  for  angles  near  90° ;  for  the  cotangent  they  are 
irregular  for  angles  near  0°. 

(2)  Logarithmic   functions. — The    principle   of    propor- 
tional parts  fails  both  for  angles  near  0°  and  angles  near 
90°.     For  the  log  sine  and  the  log  cosecant  the  differences  are 
irregular  for  angles  near  0°,  and  insensible  for  angles  near  90°. 
For  the  log  cosine  and  the  log  secant  the  differences  are  in- 
sensible for  angles  near  0°,  and  irregular  for  angles  near  90°. 
For  the  log  tangent  and  the  log  cotangent  the  differences  are 
irregular  for  angles  near  0°  and  angles  near  90°. 


EXAMPLES.  117 

It  follows,  therefore,  that  angles  near  0°  and  angles  near 
90°  cannot  be  found  with  exactness  from  their  log  trigono- 
metric functions.  These  difficulties  may  be  met  in  three 
ways. 

(1)  For  an  angle  near  0°  use  the  principle  that  the  sines 
and  tangents  of  small  angles  are  approximately  proportional 
to  the  angles  themselves.     (See  Art.  130.) 

(2)  For  an  angle  near  90°  use  the  half  angle  (Art.  99). 

(3)  In  using  the  proportional  parts,  find  two,  three,  or 
more  orders  of  differences  (Alg.,  Art.  197). 

Special  tables  are  employed  for  angles  near  the  limits  of 
the  quadrant. 

EXAMPLES. 

1.  Given  log,0  7  =  .8450980,  find  Iog10  343,  Iog10  2401,  and 
Iog10  16.807.  Ans.  2.5352940,  3.3803920,  1.2254900. 

2.  Find  the  logarithms  to  the  base  3  of  9,  81,  £,  ^,  .1,  ¥V 

Ans.  2,  4,  -  1,  -  3,  -  2,  -  4. 

3.  Find   the  value  of  Iog28,  Iog2.5,  Iog3243,  Iog5(.04), 
Iog101000,  logw  .001.  Ans.  3,  -  1,  5,  -  2,  3,  -  3. 

4.  Find  the  value  of  loga  a%  Iog6^&*,  Iog82,  log^  3,  Ioglfl010. 

.  |,  |,  £,  £,  i 


Given  Iog10  2  =  .3010300,  log,0  3  =  .4771213,  and  Iog107  = 
.8450980,  find  the  values  of  the  following  : 

5.  Iog1035,  Iog10150,  Iog10.2. 

Ans.  1.544068,  2.1760913,  1.30103. 

6.  loglo  3.5,  loglo  7.29,  log]0.  081. 

Ans.  .5440680,  .8627278,  2.9084852. 


7.    logloi,  Ioglo35,  lo 

Ans.  .3679767,  2.3856065,  .0780278. 


118  PLANE   TRIGONOMETRY. 

8.  Write  down  the  integral  part  of   the  common  loga- 
rithms of  7963,  .1,  2.61,  79.6341,  1.0006,  .00000079. 

Ans.  3,  -  1,  0,  1,  0,  -  7. 

9.  Give  the  position  of  the  first  significant  figure  in  the 
numbers  whose  logarithms  are 

2.4612310,  1.2793400,  6.1763241. 

10.  Give  the  position  of  the  first  significant  figure   in 
the   numbers  whose   logarithms   are   4.2990713,    .3040595, 
2.5860244,  3.1760913,  1.3180633,  .4980347. 

Ans.  ten  thousands,  units,  hundreds,  3rd  dec.  pi.,  1st 
dec.  pi.,  units. 

11.  Given  log  7  =  .8450980,  find  the  number  of  digits  in 
the  integral  part  of  71(),  496,  343A'*,  (V0)20,  (4.9)12,  (3.43)10. 

Ans.  9,  11,  85,  4,  9,  6. 

12.  Find  the  position  of  the  first  significant  figure  in  the 
numerical  value  of  207,   (.02)7,   (.007)2,   (3.43)^,   (.0343)8, 
(.0343)  TV 

Ans.  tenth  integral  pi.,  12th  dec.  pi.,  5th  dec.  pi.,  units, 
12th  dec.  pi.,  1st  dec.  pi. 

Show  how  to  transform 

13.  Common  logarithms  to  logarithms  with  base  2. 

Ans.  Divide  each  logarithm  by  .30103. 

14.  Logarithms  with  base  3  to  common  logarithms. 

Ans.  Multiply  each  log  by  .4771213. 

15.  Given  Iog10  2  =  .3010300,  find  Iog2 10.  3.32190. 

16.  Given  Iog10  7  =  .8450980,  find  Iog7 10.  1.183. 

17.  Given  Iog10  2  =  .3010300,  find  Iog8 10.  1.10730. 

18.  The  mantissa  of  the  log  of  85762  is  9332949;  find 
(1)  the  log  of  ^.0085762,  and  (2)  the  number  of  figures  in 
(85762) ",  when  it  is  multiplied  out. 

Ans.  (1)  1.8121177,  (2)  55. 


EXAMPLES.  119 

19.  What   are   the   characteristics  of  the  logarithms  of 
3742  to  the  bases  3,  6,  10,  and  12  respectively  ? 

Ans.  7,  4,  3,  3. 

20.  Prove  that  7  log  j-f  +  6  log  f  +  5  log  £  +  log  f  f  =  log  3. 

21.  Given  log,0  7,  find  Iog7  490.  Ans.  2  -f  -    — 

logio  7 

22.  From  5.3429  take  3.6284.  3.7145. 

23.  Divide  13.2615  by  8.  2.4076. 

24.  Prove  that  6  log  f  +  4  log  T\  +  2  log  y  =  0. 

25.  Find  log  [297  VII}*  to  the  base  3Vll.  1.8. 

Given  log  2  =  .3010300,  log  3  =  .4771213. 

26.  Find  log  216,  6480,  5400,  f. 

Ans.  2.3344539,  3.8115752,  3.7323939,  1.6478174. 


27.  Find  log  .03,  6~ 

Ans.  2.4771213,  1.7406162,  1.6365006. 

28.  Find  log  .18,  log  2.4,  logT3g, 

-4ns.  1.2552726,  .3802113,  1.2730013. 

29.  Find  log  (6.25)*,  log4V005.  .1136971,  1.45154. 

30.  Given  log  56321  =  4.7506704, 

log  56322  =  4.7506781; 
find  log  5632147.  6.7506740. 

31.  Given  log  53403  =  4.7275657, 

log  53402  =  4.  7275575; 
find  log  5340234.  6.7275603. 

32.  Given  log  56412  =  4.7513715, 

log  56413  =  4.7513792; 
find  log  564.123.  2.7513738. 


120  PLANE   TRIGONOMETRY. 

33.  Given  log  87364  =  4.9413325, 

log  87365  =  4.9413375; 
find  log  .0008736416.  Ans.  4.9413333. 

34.  Given  log  37245  =  4.5710680, 

log  37246  =  4.5710796; 
find  log  3.72456.  .5710750. 

35.  Given  log  32025  =  4.5054891, 

log  32026  =  4.5055027; 
find  log  32.025613.  1.5054974. 

36.  Given  log  65931  =  4.8190897, 

log  65932  =  4.8190962; 
find  log  .000006593171.  6.8190943. 

37.  Given  log  25819  =  4.4119394, 

log  25820  =  4.4119562; 
find  log  2.581926.  .4119438. 

38.  Given  log  23454  =  4.3702169, 

log  23453  =  4.3701984; 
find  log  23453487.  7.3702074. 

39.  Given 


log  45740  =  4.6602962, 
log  45741  =  4.6603057; 
find  the  number  whose  logarithm  is  4.6602987. 


45740.26. 


40.   Given  log  43965  =  4.6431071, 

log  43966  =  4.6431170; 
find  the  number  whose  logarithm  is  4.6431150.    .000439658. 


41.    Given  log  56891  =  4.7550436, 

log  56892  =  4. 7550512; 
find  the  number  whose  logarithm  is  .7550480. 


5.689158. 


EXAMPLES.  121 

42.  Given  log  34572  =  4.5387245, 

log  34573  =  4.5387371; 
find  the  number  whose  logarithm  is  2.5387359. 

Ans.  345.7291. 

43.  Given  log  10905  =  4.0376257, 

log  10906  =  4.0376655 ; 
find  the  number  whose  logarithm  is  3.0376371.      1090.5286. 

44.  Given  log  25725  =  4.4103554, 

log  25726  =  4.4103723 ; 
find  the  number  whose  logarithm  is  7.4103720. 

Ans.  .00000025725982. 

In  the  following  six  examples  the  student  must  take  his 
logarithms  from  the  tables. 

45.  Kequired  the  product  of  3670.257  and  12.61158,  by 
logarithms.  Ans.  46287.74. 

46.  Required  the  quotient  of  .1234567  by  54.87645,  by 
logarithms.  Ans.  .002249721. 

47.  Kequired  the  cube  of  .3180236,  by  logarithms. 

Ans.  .03216458. 

48.  Required  the  cube  root  of  ,3663265,  by  logarithms. 

Ans.  .7155216. 

49.  Required  the  eleventh  root  of  63.742.  1.45894. 

50.  Required  the  fifth  root  of  .07.  .58752. 

51.  Given  sin 42°  21'  =  .6736577, 

sin  42°  22' =.6738727; 
find  sin  42°  21' 30".  .6737652. 

52.  Given  "sin 67° 22'=  .9229865, 

sin  67°  23'  =  .9230984 ; 
find  sin  67°  22'  48".5.  .9230769. 


122  PLANE   TRIGONOMETRY. 

53.  Given  sin   7°  17'  =  .1267761, 

sin   7°  18' =  .1270646 ; 
find  sin    7°  17' 25".  Ans.  .1268963. 

54.  Given  cos  21°  27' =  .9307370, 

cos  21°  28' =  .9306306; 
find  cos  21°  27' 45".  .9306572. 

55.  Given  cos  34°  12' =  .8270806, 

cos  34°  13'  =  .8269170 ; 
find  cos  34°  12'  19".6.  .8270272. 

56.  Given  sin  41°  48'  =  .6665325, 

sin  41°  49' =  .6667493; 
find  the  angle  whose  sine  is  .6666666.  41°  48'  37". 

57.  Given  sin  73°  44' =  .9599684, 

sin  73°  45' =  .9600499; 
find  the  angle  whose  sine  is  .96.  73°  44'  23".2. 

58.  Given  cos  75°  32' =  .2498167, 

cos  75°  31 '=.2500984; 
find  the  angle  whose  cosine  is  .25.  75°  31'  21". 

59.  Given  cos  53°    7' =  .6001876, 

cos  53°    8' =  .5999549; 
find  the  angle  whose  cosine  is  .6.  53°  7'  48 ".4. 

60.  Given       log  sin  45°  16' =  9.8514969, 

log  sin  45°  17'  =  9.8516220 ; 
find  log  sin  45°  16'  30".  9.8515594. 

61.  Given       log  sin  38°  24' =  9.7931949, 

log  sin  38°  25'  =  9.7933543  ; 
find  log  sin  38°  24'  27".  9.7932666. 


EXAMPLES. 


123 


62.    Given       log  sin  32°  28' =  9.7298197, 
log  sin  32°  29'  =  9.7300182 ; 
find  log  sin  32°  28' 36".  Ans.  9.7299388. 

.63.    Given       log  sin  17°    1'  =  9.4663483. 
log  sin  17°    0'  =  9.4659353 ; 
find  log  sin  17°    0'12".  9.4660179. 

64.  Given        log  sin  26°  24' =  9.6480038. 

log  sin  26°  25'  =  9.6482582; 
find  log  sin  26°  24'  12".  9.6480547. 

65.  Given       log  cos  17°  31' =  9.9793796, 

log  cos  17°  32'  =  9.9793398 ; 
find  log  cos  17°  31'  25".2.  9.9793629. 

66.  Given       log  tan  21°  17' =  9.5905617, 

log  tan  21°  18'  =  9.5909351 ; 
find  log  tan  21°  17'  12".  9.5906364. 

67.  Given       log  tan  27°  26'  =  9.7152419, . 

log  tan  27°  27'  =  9.7155508 ; 
find  log  tan  27°  26' 42".  9.7154581. 

68.  Given       log  cot  72°  15'  =  9.5052891, 

log  cot  72°  16'  =  9.5048538 ; 
find  log  cot  72°  15'  35".  9.5050352. 

69.  Given       log  cot  36°  18'  =  10.1339650, 

log  cot  36°  19'  =  10.1337003 ; 
find  log  cot  36°  18' 20".  10.1338768. 

70.  Given        log  cot  51°  17' =  9.9039733, 

log  cot  51°  18'  =  9.9037144 ; 
find  .  log  cot  51°  17'  32".  9.9038352. 


124  PLANE  TRIGONOMETRY. 

71.  Given       log  sin  16°  19'  =  9.4486227, 

log  sin  16°  20'  =  9.4490540 ; 
find  the  angle  whose  log  sine  is  9.4488105. 

Ans.  16°  19' 26". 

72.  Given       log  sin   6°  53'        =9.0786310, 

log  sin    6°  53' 10"  =9.0788054; 
find  the  angle  whose  log  sine  is  9.0787743.  6°  53'  8". 

73.  Given      log  cos  22°  28'  20"=  9.9657025, 

log  cos  22°  28'  10"=  9.9657112; 
find  the  angle  whose  log  cosine  is  9.9657056.  22°  28'  16". 

In  the  following  examples  the  tables  are  to  be  used : 

74.  Find  log  tan  55°  37' 53".  Ans.  10.1650011. 
70.    Find  log  sin  73°  20'  15". 7.  9.9813707. 

76.  Find  log  cos  55°  11'  12".  9.7565636. 

77.  Find  log  tan  16°    0'27".  9.4577109. 

78.  Find  log  sec  16°    0'27".  10.0171747. 

79.  Find  the  angle  whose  log  cosine  is  9.9713383. 

Ans.  20°  35' 16". 

80.  Find  the  angle  whose  log  cosine  is  9.9165646. 

Ans.  34°  23' 25". 

81.  Find  log  cos  34°  24'  26".  9.9164762. 

82.  Find  log  cos  37°  19'  47".  9.9004540. 

83.  Find  log  sin  37°  19'  47".  9.7827599. 

84.  Find  log  tan  37°  19'  47".  9.8823059. 

85.  Find  log  sin  32°  18'  24".6.  9.7279096. 

86.  Findlogcos32°18'24".6.  9.9269585. 

87.  Find  log  tan  32°  18'  24".6.  9.8009511. 


EXAMPLES.  125 

Prove  the  following  by  the  use  of  logarithms  : 


^512  x  ^.00003075  =  .000232432. 
SX80  -=-  -v/.OOOOOOl 


90.    W(2002)""j<Il( «1)^2  =  218403ooooO. 
\          1001  v  2002 


126  PLANE   TRIGONOMETRY. 


CHAPTER   V. 
SOLUTION   OF  TKIGONOMETKIO  EQUATIONS, 

82.  A  Trigonometric  Equation  is  an  equation  in  which 
the  unknown  quantities  involve  trigonometric  functions. 

The  solution  of  a  trigonometric  equation  is  the  process  of 
finding  the  values  of  the  unknown  quantity  which  satisfy  the 
equation.  As  in  Algebra,  we  may  have  two  or  more  simul- 
taneous equations,  the  number  of  angles  involved  being 
equal  to  the  number  of  equations. 

EXAMPLES. 


1.  Solve  sin  0  =  — 

2 

This  is  a  trigonometric  equation.     To  solve  it  we  must 
find  some  angle  whose  sine  is  —     We  know  that  sin 30°=  r- 

Therefore,  if  30°  be  put  for  0,  the  equation  is  satisfied. 
.-.  0  =  30°  is  a  solution  of  the  equation. 

.-.  0  =  ?i,r +  (-!)"£ (Art.  38) 

0 

2.  Solve  cos  0  + sec  0  =  -. 

The  usual  method  of  solution  is  to  express  all  the  func- 
tions in  terms  of  one  of  them. 

Thus,  we  put for  sec0,  and  get 

COS0 


cos0     2 


TRIGONOMETRIC  EQUATIONS.  127 

This  is  an  equation  in  which  0,  and  therefore  cos0,  is 
unknown.  We  proceed  to  solve  the  equation  algebraically 
just  as  we  should  if  x  occupied  the  place  of  cos  6,  thus  : 

cos2  0  -    cos  B  =  -1. 


S- 

The  value  2  is  inadmissible,  for  there  is  no  angle  whose 
cosine  is  numerically  greater  than  1  (Art.  21). 

.-.  0080==-*- 
& 

But  cos  60°  =  1. 

.-.  cos  0  —  cos  60°. 
Therefore  one  value  of  0  which  satisfies  the  equation  is  60°. 

3.  Solve    cosec  6  -  cot20  +  1  =  0. 

We  have     cosec  0  -  (cosec2  0  —  1  )  +  1  =  0  .     .     (Art.  23) 
cosec20  —  cosec  0  =  2. 

,.  cosec0=]±| 

=  2  or  -  1. 
But  cosec  30°  =  2. 

.  •.  cosec  0  =  cosec  30°. 
Therefore  30°  is  one  value  of  0  which  satisfies  the  equation  . 

Find  a  value  of  0  which  will  satisfy  the  following  equa- 
tions : 

4.  cos  0  =  cos  20.  Ans.  ITT. 

5.  2  cos  0  =  sec  0.  45°. 


128  PLANE   TRIGONOMETRY. 

6.  4  sin  0  —  3  cosec  0  =  0.  Ans.  60°. 

7.  4  cos  0  =  3  sec  0.  30°. 

8.  3sin0-2cos20  =  0.  30°. 

9.  V2  sin  0  =  tan  0.  0°  or  45°. 

10.  tan  0  =  3  cot  0.  60°. 

11.  tan  0  + Scot  0  =  4.  45°. 

12.  cos  0  + cos  30  + cos  50  =  0.  £,  ^ 

^      o 

13.  sin  (0  -<£)  =  %,  cos  (0  +  <£)  =  0.          0  =  60°,  <£  =  30°. 

83.    Solve  the  equations 

m  sin  (f>  =  a (1) 

m  cos  <£  =  & (2) 

where  a  and  b  are  given,  and  the  values  of  m  and  <£  are 
required. 

Dividing  (1)  by  (2),  we  get 

tan  4  =  -, 
b 

which  gives  two  values  of  <£,  differing  by  180°,  and  there- 
fore two  values  of  m  also  from  either  of  the  equations 

a  b 

m  =  -    —  =  —   — 
sin  </>      cos  <£ 

The  two  values  of  m  will  be  equal  numerically  with 
opposite  signs. 

In  practice,  m  is  almost  always  positive  by  the  conditions 
of  the  problem.  Accordingly,  sin  <£  has  the  sign  of  a,  and 
cos  <£  the  sign  of  b}  and  hence  <£  must  be  taken  in  the  quail- 
rant  denoted  by  these  signs.  These  cases  may  be  considered 
as  follows : 

(1)  Sin  <£  and  cos  </>  both  positive.  This  requires  that  the 
angle  <£  be  taken  in  the  first  quadrant,  because  sin  <£  and 
cos  <£  are  both  positive  in  no  other  quadrant. 


TRIGONOMETRIC  EQUATIONS.  129 

(2)  Sin  <f>  positive  and  cos  <£  negative.    This  requires  that  $ 
be  taken  in  the  second  quadrant,  because  only  in  this  quad- 
rant is  sin  <f>  positive  and  cos  <£  negative  for  the  same  angle. 

(3)  Sin  (f>  and  cos  <£  both  negative.     This  requires  that  <f> 
be  taken  in  the  third  quadrant,  because  only  in  this  quad- 
rant are  sin  <f>  and  cos  <j>  both  negative  for  the  same  angle. 

(4)  Sin  <f>  negative  and  cos  <f>  positive.     This  requires  that 
<£  be  taken  in  the  fourth  quadrant,  because  only  in  this 
quadrant  is  sin  <f>  negative  and  cos  <f>  positive  for  the  same 
angle. 

Ex.  1.     Solve  the  equations  msin  <£  =  332.76,  and  mcos  <£ 
=  290.08,  for  m  and  <£. 

log  m  sin  <£  =  2.52213 
log  m  cos  <£  =  2.46252 

log  tan  <£  =  0.05961 
.-.  <£  =  48°55'.2. 

log  m  sin  <f>  =  2.52213 

log  sin  4  =  9.87725 

log  m  =  2.64488 
.-.  m  =  441.45. 

Ex.  2.    Solve  m  sin  <£  =  -  72.631,  and  m  cos  $  =  38.412. 

Aiis.  <t>  =  117°  52'.3,  m  =  -  82.164. 

84.   Solve  the  equation 

a  sin  <£  +  &  cos  </>  =  c (1) 

a,  6,  and  c  being  given,  and  <£  required. 

Find  in  the  tables  the  angle  whose  tangent  is  - ;  let  it 
be/?. 

Then  -  =  tan  /?,  and  (1)  becomes 

a  (sin  <f>  +  tan  ft  cos  <£)  =  c ; 

/sin  <£  cos  (3  +  cos  <£  sin  J3\ 

or         ci  ( ]  —  c  j 

V  cos  ft  ) 

or  sin(</>-f ^)  =  -cos^  =  -siny8    ....     (2) 

a  o 


130  PLANE  TRIGONOMETRY. 

There  will  be  two  solutions  from  the  two  values  of  <£  -f-  ft 
given  in  (2). 

Find  from  the  tables  the  value  of  cos  ft.     Next  find  from 

the   tables   the  magnitude  of  the  angle  a  whose  sine  =  - 
cos  /3,  and  we  get 

sin  (<£  +  /?)  =  sin  a, 

.-.$  +  £  =  wir  +  (-!)"«     .     .    (Art.  38) 
nir        -l' 


where  n  is  zero  or  any  positive  or  negative  integer. 

In  order  that  the  solution  may  be  possible,  it  is  necessary 

to  have  -  cos  J3  =,  or  <  1. 
a 

NOTE.—  This  example  might  have  been  solved  by  squaring  both  sides  of  the 
equation;  but  in  solving  trigonometric  equations,  it  is  important,  if  possible,  to 
avoid  squaring  both  sides  of  the  equation. 

Thus,  solve  COB  9  =  k  sin  0      ...........     (3) 

If  we  square  both  sides  we  get 

cos2  0  =  Jfl  sin2  0  =  fc2(i  -  cos2  0)  . 


(4) 


Now  if  a  be  the  least  angle  whose  cosine  =  —    —  ,  we  get  from  (4) 

*S\  +  k2 
0  =  mr±a     ...........     (5) 

But  (3)  may  be  written  cot  9  =  k. 

.'.  e  =  nn  +  a      ...........      (6) 

(6)  is  the  complete  solution  of  the  given  equation  (3),  while  (5)  is  the  solution  of 
both  cos  0  =  k  sin  0,  and  also  of  cos  0  =  —  k  sin  0.  Therefore  by  squaring  both  mem- 
bers of  an  equation  we  obtain  solutions  which  do  not  belong  to  the  given  equation. 

EXAMPLES. 

1.  Solve  0.7466898  sin  <£  -  1.0498  cos  </>  =  -  0.431689, 
when  <£  <  180°. 

log  b  =  0.02112-* 

log  a  =  1.87314 
log  tan  £  =  0.14798  - 
.-.  ft  =  125°  25'  20". 

*  The  minus  sign  is  written  thus  to  denote  that  it  belongs  to  the  natural  number 
and  does  not  affect  the  logarithm.  Sometimes  the  letter  n  is  written  instead  of  the 
minus  sign,  to  denote  the  same  thing. 


TRIGONOMETRIC  EQUATIONS.  131 

log  sin  ft  =  9.91111 

log  c=  1.63517  - 
colog&  =  9.97888- 

log  sin  (<£  +  £)=  9.52516+ 

...  <£  +  ft  =  19°  34'  40"  or  160°  25'  20". 
.-.  <£  =  -  105°  50'  40"  or  35°  0'  0". 

2.  Solve  -  23.8  sin  <f>  +  19.3  cos  0  =  17.5(<£  <  180°). 

Ans.  <f>  =  4°  12'.7  or  -  106°  7'.9. 

3.  Solve  2  sin  0  +  2  cos  0  =  V2.  Ans.  -  -+mr  +  (-l)n^- 

4  6 


4.  . 

o  b 

5.  «      sin^-cos^  =  l.  -  +  W7r  +  (-l)n^. 

6.  "      V3  sin  0  -  cos  6  =  V2.        ^-TT  +  WTT  +  (—  I)ni7r. 

85.   /Sofae  ^/ie  equation 

sin  (a  +  »)  =  msinx  ......     (1) 

in  which  a  and  m  are  given. 
From  (1)  we  have 

sin  (a  +  #)  +-  sin  a;  _  m  +•  1  /2\ 

sin  (a  -f-  x)  —  sin  x      m  —  1 


(Art.  46) 
.     .     (3) 


tan 


which  determines  x  +  ^-a,  and  therefore  a?. 

If  we  introduce  an  auxiliary  angle,   the  calculation  of 
equation  (3)  is  facilitated. 


132  PLANE   TRIGONOMETRY. 

Thus,  let  m  =  tan<£;  then  we  have  by  [(14)  of  Art.  61] 

mill  =  tan^  +  l  =  co 
ra-1      tan<£-l 

which  in  (3)  gives 

tan  (x  +  £}  =  cot  (<£  -  45°)  tan  £  a     .     (4) 
\        */ 

This,  with  tan  <£  =  m, 

gives  the  logarithmic  solution. 

The  logarithmic  solution  of  the  equation 

sin  (a  —  x)  =  m  sin  x 
is  found  in  the  same  manner  to  be 
tan  <f>  =  m, 

and  tan  fx  - 1^  =  cot  (<£  +  45°)  tan  £ 

which  the  student  may  show. 
Example.  —  Solve 

sin  (106°+  a;)  =  -  1.263  sin  a  (a  <  180°). 
log  tan<£  =  logm  =  log  (-  1.263)  =  0.10140  -. 

.-.  <£  =  128°  22'.3. 

4>_45°=    83°22'.3;       log  cot  (<£  -  45°)  =    9.06523 
i«=    53°   O'.O,  log tan^a  =  10.12289 


-]  =    9.18812 
V       2/ 

|  =  8°  46'.0  or  188°  46'. 
.-.  x  =  -44°  14'  or  135°  46'. 

86.   Solve  the  equation 

tan(tt-f-#)  =  m  tana; (1) 

in  which  a  and  m  are  given. 


TRIGONOMETRIC  EQUATIONS.  133 

From  (1)  we  have 

tan  («  +  x)  +  tan  x  _  m  +  1 
tan  (a  +  x)  —  tan  x      m  —  1 

=  si"("+2*)    [(21)  of  Art.  61] 
sma 

fffll        -..,_    1 

=  cot(<£  -  45°)  sin  a   .     (Art.  85) 
where  tan  <f>  =  m. 

Example.  —  Solve  tan (23°  16'+  x)  =  .296  tana;, 
log  tan  4>  =  logm  ==  log(.296)  =  1.47129. 

.-.  <£  =  16°  29'.3. 

<£  -  45°=  -  28°  30'.7 ;      log  cot(<£  -  45°)  =  10.26502- 
«  =  23°16'.l,  log  sin  a  =    9.59661 

logsin(a  +  2z)  =    9.86163- 
a  +  2  x  =  226°  38'.9  or  313°  21M. 
.-.  x  =  101°  41'.5  or  145°2'.6. 

87.   Solve  the  equation 

tan(a  +  a;).tanx  =  m (1) 

in  which  a  and  m  are  given. 

From  (1)  we  have 
1  +  tan  (a  +  a;)  tan  x  _  1  +  m 
1  —  tan  (a  +  x)  tan  x      1  —  m 

= ^V        .     .    (Ex.  4  of  Art.  47) 

cos  (a +  2  a) 

-| 

:  — ^ COS  « 

1  +  m 

=  tan  (45°-  <£)  cos  a     [(16)  of  Art.  61] 
where  tan  <j>  =  m. 


184  PLANE  TRIGONOMETRY. 

Example.  —  Solve  tan  (65°  +  a;)  tana;  =  1.5196  (a  <  180°). 
log  tan  <£  =  log  w  =  0.18173. 

.-.  <£  =  50°39'9"; 

45°  _  <£  =  _  11°  39'  9"  ;       log  tan(450-<£)  -  9.31434- 
a  =  65°  0'  0"  ;  log  cos  a  =  9.62595 

log  cos  (a  +  2  x)  =  8.94029- 
a  +  2«  =  95°or  265°. 

.-.  2  x  =  30°  or  200°. 
/.     x  =  15°  or  100°. 

88.   Solve  the  equations 

m  sin  (6  +  x)  =  a    ..........     (1) 

m  sin  (<£  -f  x)  =  b  ..........     (2) 

for  m  and  x,  the  other  four  quantities,  0,  <j>,  a,  b,  being 
known. 

Expanding  (1)  and  (2)  by  (Art.  44),  we  get 

m  sin  0  cos  x  +  m  cos  0  sin  x  =  a    .....     (3) 
m  sin  (f>  cos  x  -\-  m  cos  <£  sin  x  =  b  .....     (4) 

Multiplying  (3)  by  sin  <£  and  (4)  by  sin  0,  and  subtracting 
the  latter  from  the  former,  we  have 

m  sin  x  (sin  <£  cos  9  —  cos  <j>  sin  0)  =  a  sin  <£  —  6  sin  0. 

......     (5) 


To  find  the  value  of  mcoso;,  multiply  (3)  and  (4)  by 
cos  <£  and  cos  0,  respectively,  and  subtract  the  former  from 
the  latter.  Thus 

m  cos  a?  (sin  <£  cos  0  —  cos  <£  sin  0)  =  b  cos  6  —  a  cos  <£. 


(6) 


Having  obtained  the  values  of  m  sin  x  and  m  cos  x  from 
(5)  and  (6),  m  and  a;  can  be  calculated  by  Art.  83. 


TRIGONOMETRIC  EQUATIONS.  135 

EXAMPLES. 

1.    Solve   m  cos  (6  +  x)  =  a,   and   m  sin  (  <£  +  a; )  =6,   for 
m  sin  z  and  »  cos  x.  ^   m  gin  ^  =  6_coe£ 

cos (# - 


2.    Solve    m  cos  (0  +  cc)  =  a,    and    mcos  (<£  —  x)=  b,    for 
m  sin  x  and  m  cos  x.  A  b  cos  0  —  a  cos  < 


sin  (0  +  *) 


sin  (0  -f 
89.   /SWve  ^/ie  equation 


m      .......     (1) 

a;  sin  a  —  y  cos  a  =  n       .......     (2) 

for  x  and  y. 

Multiplying  (1)  by  cos  a  and  (2)  by  since,  and  adding, 
we  get 

x  =  m  cos  a  +  w  sin  «. 

To  find  the  value  of  y,  multiply  (1)  by  sin  a  and  (2)  by 
cos  a,  and  subtract  the  latter  from  the  former.  Thus 

y  =  m  sin  a  —  n  cos  a. 

Example.  —  Solve 

x  sin  a  +  y  cos  a  =  a, 
x  cos  a  —  y  sin  a  =  b. 

90.   To  adapt  Formulae  to  Logarithmic  Computation.— 

As  calculations  are  performed  principally  by  means  of 
logarithms,  and  as  we  are  not  able  by  logarithms  directly* 
to  add  and  subtract  quantities,  it  becomes  necessary  to 
know  how  to  transform  sums  and  differences  into  products 

*  Addition  and  Subtraction  Tables  are  published,  by  means  of  which  the  logarithm 
of  the  sum  or  difference  of  two  numbers  may  be  obtained.  (See  Tafeln  der  Addi- 
tions, und  Subtractions,  Logarithmen  flir  sieben  Stellen,  von  J.  Zech,  Berlin.) 


136  PLANE  TRIGONOMETRY. 

and  quotients.  An  expression  in  the  form  of  a  product  or 
quotient  is  said  to  be  adapted  to  logarithmic  computation. 

An  angle,  introduced  into  an  expression  in  order  to  adapt 
it  to  logarithmic  computation,  is  called  a  Subsidiary  Angle. 
Such  an  angle  was  introduced  into  each  of  the  Arts.  84,  85, 
86,  and  87. 

The  following  are  further  examples  of  the  use  of  sub- 
sidiary angles : 

1.  Transform  a  cos  6  ±  b  sin  0  into  a  product,  so  as  to 
adapt  it  to  logarithmic  computation. 

Put  -  =  tan  d> ;  *  thus 
a 

/  T  S 

a  cos  6  ±  b  sin  6  =  a  (  cos  0  ±  -  sin  0 


V 

=  a  (cos  B  ±  tan  <£  sin  0) 
—     a     cos  (0  T  <£)• 

COS<£ 

2.  Similarly, 

a  sin  0  ±  6  cos  0  =  — ^—  sin  (0  ±  <£). 

3.  Transform  a  ±  b  into  a  product, 

a  +  &  =  a/l-|--)  =  a(l4-  tan2  <f>)  =  a  sec2  <£, 

if  -  =  tan2  <f>. 


a 


a  —  b  =  a(  1 )  = 

V     a/ 


if  -  =  sin24>. 

a 

*  The  fundamental  formulae  cos  (a;  ±  #)  and  sin  (x  +  #)  (Art.  42)  afford  examples  of 
one  term  equal  to  the  sum  or  difference  of  two  terms ;  hence  we  may  transform  an 
expression  a  cos  0  ±6  sin  0  into  an  equivalent  product,  by  conforming  it  to  the  for- 
mulae just  mentioned. 

Thus,  comparing  the  identity,  m  cos  <f>  cos  9  ±  m  sin  <f>  sin  B  =  m  cos  (</>  T  0)  or  m  COB 
),  with  a  cos  0±  6  sin  0,  we  will  have  a  cos  0+  &sin0  =  mcoB(0  t  <f>)  if  we  assume 

and  6  =  m  sin  <f> ;  i.e.  (Art.  83),  if  tan</>  =  -  and  m=     a 
See  Art.  84.  °  cos* 


ADAPTATION   TO  LOGARITHMIC  COMPUTATION.    137 

.-.  log  (a  +  6)  =  log  a +  2  log  sec  <£; 
and  log  (a  —  6)  =  loga-f21ogcos<£. 

4.    Transform  1239.3  sin  0  -  724.6  cos  6  to  a  product. 

log  6  =  log  (-  724.6)  =  2.86010- 
loga  =  log  (1239.3)  =  3.09318 

log  tan  <£  =  9. 76692- 
.-.  <£  =  -30°18'.8. 

log  a  =  3.09318 
log  cos  <£  =  9.93615 

log  _^_  =  3.15703 


—  =  1435.6. 


.-.  1239.3  sin  0  -  724.6  cos  0  =  1435.6  sin  (0  -  30°  18'.8)  . 

91.   Solve  the  equations 

r  cos  <£  cos  0  =  a      ........     (1) 

(2) 

(3) 
for  r,  <£,  and  0. 

Dividing  (2)  by  (1),  we  have 

tan  0=5, 

a 
from  which  we  obtain  0. 

From  (1)  and  (2)  we  have 

.          .     (4) 


.     . 

COS0        S1110 

from  which  we  obtain  rcos  <£. 

From  (3)  and  (4)  we  obtain  r  and  <£  (Art.  83). 


138  PLANE  TRIGONOMETRY. 

EXAMPLES. 

1.    Solve  r  cos  0  cos  0  =  — 53.953, 

r  cos  </>  sin  0  =  197.207, 

rsin<£  =  -  39.062, 
for  r,  0,  0. 

log  6  =  2.29493 


log  r  cos  <£  =  2.31060 
log  cos  <f>  =  9.99221 
logr  =  2.31839 
.-.  r  =  208.16. 


log  a  =  1.73201- 
log  tan  0  =  0.56292  - 
.-.  0  =  105°18'.0. 
log  sin  0  =  9.98433 
log  r  cos  0  =  2.31060 
log  r  sin  0  =  1.59175- 

log  tan  0  =  9.28115- 

92.  Trigonometric  Elimination.  —  Several  simultaneous 
equations  may  be  given,  as  in  Algebra,  by  the  combination 
of  which  certain  quantities  may  be  eliminated,  and  a  result 
obtained  involving  the  remaining  quantities. 

Trigonometric  elimination  occurs  chiefly  in  the  applica- 
tion of  Trigonometry  to  the  higher  branches  of  Mathematics, 
as,  for  example,  in  Physical  Astronomy,  Mechanics,  Analytic 
Geometry,  etc.  As  no  special  rules  can  be  given,  we  illus- 
trate the  process  by  a  few  examples. 

EXAMPLES. 
1.    Eliminate  <£  from  the  equations 

x  =  a  cos  <£,  y  =  b  sin  <£. 
From  the  given  equations  we  have 

-  =  cos  <£,  ^  =  sin  <£, 
a  b 

which  in  cos2<£  -}-  sin2</>  =  1, 

gives  +  y=l. 


TMIGONOMETEIC  ELIMINATIONS.  139 

2.   Eliminate  <f>  from  the  equations 

a  cos  <f>  +  b  sin  <£  =  c, 

6  cos  <£  -f-  c  sin  <£  =  a. 
Solving  these  equations  for  sin  <£  and  cos  <£,  we  have 

•     ,_bc  —  a2 
U^-^I~^ 


--  - 
ac  —  b2 

which  in  cos2  <£  +  sin2  <£  =  1, 

gives       (be  -  a2)2  +  (c2  -  ab)2  =  (ac  -  b2)2. 

3.   Eliminate  <f>  from  the  equations 

y  cos  <£  —  x  sin  </>  =  a  cos  2  <£, 
y  sin  <^>  +  x  cos  <^>  =  2  a  sin  2  <£. 

Solve  for  x  and  y,  then  add  and  subtract,  and  we  get 


-  ?/)2  =  a2(l  -  si 


4.   Eliminate  a  and  ft  from  the  equations 

a  =  sin  a  cos  ft  sin  6  -f-  cos  a  cos  0  .  .  .  (1) 
b  =  sin  a  cos  /2  cos  0  —  cos  a  sin  0  .  .  .  (2) 
c  =  sin  a  sin/?  sin  0 (3) 

Squaring  (1)  and  (2),  and  adding,  we  get 

2     i     £,2  *2  2Oi  2  //l\ 

— — -  =  sin2«sin2/? •     (5) 

sin20 


140  PLANE  TRIGONOMETRY. 

Adding  (4)  and  (5),  we  have 

a2  +  b2  +  --^—  =  1. 
sin2  0 

5.  Eliminate  <f>  from  the  equations 

a  sin  <j>  +  6  cos  <j>  =  c, 

acos<£  —  bsiu<J>  =  d.        Ans.  a2  +  &2  <=  c2  +  c?2. 

6.  Eliminate  0  from  the  equations 

m  =  cosec  6  —  sin  0, 

n  =  sec  0  —  cos  0.  mV  (m*  +n%)  =  1. 

7.  Eliminate  0  and  <£  from  the  equations 

sin  6  -j-  sin  <£  =  a, 
cos  0  +  cos  <£  =  b, 
cos(0  -  0)  =  c.  a2  +  V2  -  2c  =  2. 

8.  Eliminate  x  and  ?/  from  the  equations 

tan  x  -f-  tan  37  =  a, 
cot  x  +  cot  y  =  6, 

a  +  y  =  c.  cot  c  =  -  —  -. 

a      b 

9.  Eliminate  <f>  from  the  equations 

x  =  cos  2  <£  +  cos  <£, 
y  =  sin  2  <£  +  sin  <£. 


EXAMPLES. 

Solve  the  following  equations  : 

1.  tan  0  +  cot  0  =  2.  Ans.  45°. 

2.  2  sin2  0  +  V2  cos  0  =  2.  90°,  or  45°. 

3.  3tan20-4sin20  =  l.  45°. 

4.  2sin20  +  V2sin0  =  L>.  45°. 


EXAMPLES.  141 

5.  cos2  0-V3  cos  0  +  f  =  0.  Ans.  30°. 

6.  sin50  =  16  sin5  0.  TITT,  or  mr  ±  -• 

6 

7.  sin  9  0  —  sin  6  =  sin  4  0.  Jn?r,  or  f  n?r  ±  -^. 

To 

8.  2  sin  0  =  tan  0.  mr,  or  2  TITT  ±  -• 

o 

9.  6cot20-4cos20  =  l.  7i7r±-. 

3 

10.  tan  0  +  tan  (0  -45°)  =  2.  mr  ±  -. 

o 

11.  cos  0  4-  V3  sin  0  =  V2.  2  WTT  ±  -• 

3 

12.  tan  (6  +  45°)  =  1  +  sin  2  0.  ww  —  -  ,  or  7i7r. 


13.  (cot0-tan0)2(2  +  V3)  =  4(2 

14.  cosec  0  cot  0  =  2  V3.  2nir±-- 

6 

15.  cosec  0  +  cot  (9  =  V3. 


16.  sin-  =  cosec  0  —  cot  0.  2?*7r. 

17.  sin  50  cos  30  =  sin  90  cos  70.  J^-WTT  +  (-  1)|- 


18.  siri20  +  cos22<9  =  f.  mr  ±  —  ,  or  mr  ±  T%  TT. 

19.  V3sin0-COS0  =  V2.  TITT  +  ^  +  (-  l)nr 

6  4 

20.  tan  0  +  cot  0  =  4.  WIT  + 

21.  si 

22.  Solve  m  sin  <#>  =  1.^9743,  and  m  cos  <#>  =  6.0024. 


142  PLANE  TRIGONOMETRY. 

23.  Solve  m  sin  <£=  -0.3076258,  and  m  cos  0=0.4278735. 
(m  positive.)  Ans.  <£  =  324°  17'  6".6,  ?>i  =  0.52698. 

24.  Solve  m  sin  0  =  0.08219,  and  m  cos  0  =  0.1288. 

25.  Solve  m  sin  <£  =  194.683,  and  m  cos  <£  =  8460.7. 

26.  If   a  sin  0  4-  6  cos  0  =  c,    and   a  cos  0  +  6  sin  6  =  c  sin0 
cos  0,  show  that  sin  2  <9(c'2  -  a2  -  62)  =  2  a&. 

Solve  the  following  equations  : 

27.  V2  sin  0  +  V2  cos  0  =  V3.   ^bis.  —  -  +  mr  +  (—  l)n-. 

28.  2  sin  a;  +  5  cos  a;  =  2.     Sug.  [2.5  =  tan  68°  12']. 

Ans.  x  =  -6S°  12'  -f  71  180°  +  (  -  1)"(21°  48'). 

29.  3  cos  a  —  8  sin  a;  =  3.     Sug.  [2.6  =  tan  69°  26'  30"]. 

Ans.  x  =  -  69°  26'  30"  +  2  n  180°  ±  (69°  26'  30"). 

30.  4  sin  x  —  15  cos  x  =  4.     Sug.  [3.75  =  tan  75°  4']. 

.4HS.  x  =  75°  4'  +  n  180°  +  (-  1)H(14°  56'). 

31.  cos  (a  -)-«)=  sin  (a  -f  x)  -f-  V2cos^8. 

^l?is.  ic  =  —  «  —  -  +  2  n?r  ±  /?. 
4 

32.  cos  0  +  cos  3  0  H-  cos  5  0  =  0. 

Ans.   l(2n  +  I)TT,   or  1(3  ?i  ±  I)TT. 

33.  sin  5  0  =  sin  3  0  +  sin  0  =  3  -  4  sin2  0. 


±    ,  or  |(2?i  +  I)TT. 
3 


34. 

Ans. 

35. 


or  cos"1- 


EXAMPLES.  143 

36.    Solve 

m  sin(0  -\-x)  =  a  cos  /?,  and  m  cos  (0  —  #)  =  ei  sin  /?, 
for  msina;  and  racosx.     (Art.  67.) 

Ans. 


cos  20 

a  sin  (5  —  0) 
m  cos  a;  =  —  —  ^- 

cos  2  (9 

37.  Solve  m  cos  (0  +  <£)  =  3.  79,  and  ra  cos  (0  —  <£)  =  2.06, 
for  m  and  0,  when  <£  =  3J°  27'.4.     (Art.  67.) 

38.  Solve  r  cos  <£  cos  0  =  1.271, 

r  cos  4>  sin  0  =  -  0.981, 

r  sin  4»  =  0.890, 
for  r,  <#>,  0.     (Art.  70.) 

39.  Solve  rcos<£cos0  =  —  2, 

r  cos  <£  sin  0  =  -f  3, 

r  sin  <£  =  —  4, 
for  r,  <£,  0. 

40.  Solve  r  sin  <£  sin  0  =  19.765, 

r  sin  <£  cos  0  =  -  7.192, 

r  cos  4>  =  12.124, 
for  r,  <£,  (9. 


41.  Solve  cos(2aj  +  3y)  =  J,  cos(3x  +  2?/)  =  |V3. 

^TiS.    i»=|W7r±T^7r±  LTT,   y  =  |7l7r  ±  |TT  ±  ^ 

42.  Solve  cos  3  0  +  cos  50  +V2  (cos  0  +  sin0)  cos  0  =  0. 

Ans.  4(9±0  =  2?i7r±     TT    or 


43.  Solve  c 

Ans.  sin  0  =  0,  or  tan0  =  —  l±v/2. 

44.  Solve  3  sin  0  -f  cos  0  =  2  x,  sin  0  -f  2  cos  0  =  x. 

Ans.  0  =  71°  34',  aj  = 


144  PLANE   TRIGONOMETRY. 

45.  Solve  1.268  sin  <£  =  0.948  +  m  sin  (25°  27'.2), 

1.268  cos  4»  =  0.281  +  m  cos  (25°  27'.2). 

Ans.  </>  =  60°  53'.8,  m  =  0.372. 

46.  Transform    at  +  y*  +  z4  —  2  y*z*  —  2  zW  —  2  a?y*  into  a 
product.        ^.7?s.  —  (#+?/+2)  (y+z  —  x)  (z+x—  y)  (x-\-y—  z). 

47.  Eliminate  0  from  the  equations 

m  sin  2  0  =  n  sin  0,  p  cos  2  0  ==  g  cos  0. 


m 


2        2 


48.   Eliminate  0  and  <j>  from  the  equations 

x  =  a  cosm  0  cosm  <f>,  y  =  b  cos™  6  sinw  <£,  2  =  c  sin" 


49.  Eliminate  6  from  the  equations 

a  sin  0  +  6  cos  6  =  h,  a  cos  0  —  b  sin  0  =  k. 

Ans.  a2  +  b2  =  h2  +  fc2. 

50.  Eliminate  0  from  the  equations 

a  tan  0  +  6  sec  B  =  c,  a'  cot  0  +  6'  cosec  0  =  c'. 

Ans.  (a'b  +  cb')2  +  (06'  +  c'&)2  =  (ccf  -  aa')2. 

51.  Eliminate  0  from  the  equations 
x  =  2  a  cos  0  cos  2  0  —  a  cos  0, 

y  =  2a  cos  0  sin  2  0  —  a  sin  0.  ^.ns.  ic2  +  y2  =  a2. 

52.  Eliminate  0  from  the  equations 

x  =  a  cos  0  +  6  cos  2  0,  and  y  =  a  sin  0  +  6  sin  2  0. 

Ans.  a2  [(x  +  6)2  +  /]  =  [x2  +  7/2  -  62]2. 

53.  Eliminate  a  and  /3  from  the  equations 
6  +  c  cos  a  =  u  cos  (a  —  0), 

&  +  CCOS/?  =  MCOS  (ft  —  0),  a  —  /?  =  2<£; 
and  show  that 

u2  —  2  we  cos  0  +  c2  =  62  sec2  <£. 


EXAMPLES  IN  ELIMINATION.  145 

54.   Eliminate  0  and  </>  from  the  equations 

x  cos  6  +  y  sin  0  =  a,  b  sin  (0  -f-  <£)  =  a  sin  <£, 
a  cos  (0  +  2cf>)  -  i/  sin  (0  +  2<£)  =  a. 


55.  Eliminate  0  from  the  equations 

#_  sec20  —  cos20 
a     sec20  +  cos20' 

^  =  sec20  +  cos20. 
y 

56.  Eliminate  0  from  the  equations 

(a  +  6)  tan  (0  -  <f>)  =  (a  -  b)  tan  (0  +  <£), 


57.    Eliminate  ^  from  the  equations 


58.   Eliminate  ^  and  <^>  from  the  equations 

a2  cos2  0  —  b2  cos2  (f>  =  c2,  acosQ  +  b  cos  <£  =  r, 
a  tan  &  =  b  tan  <£. 


59.  Eliminate  <^>  from  the  equations 
n  sin  0  —  m  cos  0  =  2  m  sin  </>, 

n  sin  2  0  —  m  cos  2  <£  =  n. 

Ans.   (nsin#-f-  mcos#)2  =  2m  (m  +  n). 

60.  Eliminate  a  from  the  equations 
x  tan  (a  —  /?)  =  y  tan  («  +  /?), 

(#  —  y)  cos  2  a  +  («  +  y)  cos2fi  =  z. 
Ans. 


146  PLANE  TRIGONOMETRY. 


CHAPTER  VI. 

EELATIONS    BETWEEN    THE    SIDES    OP  A  TKIANGLE 
AND  THE  FUNCTIONS   OF  ITS   ANGLES, 

93.  Formulae.  —  In  this  chapter  we  shall  deduce  formulae 
which  express  certain  relations  between  the  sides  of  a  tri- 
angle and  the  functions  of  its  angles.     These  relations  will 
be  applied  in  the  next  chapter  to  the  solution  of  triangles. 
One  of  the  principal  objects  of  Trigonometry,  as  its  name 
implies  (Art.  1),  is  to  establish  certain  relations  between 
the  sides  and  angles  of  triangles,  so  that  when  some  of 
these  are  known  the  rest  may  be  determined. 

RIGHT   TRIANGLES. 

94.  Let  ABC  be  a  triangle,  right-angled  at  C.     Denote 
the  angles  of  the  triangle  by  the  let- 
ters A,  B,  C,  and  the  lengths  of  the 

sides  respectively  opposite  these  an- 
gles, by  the  letters  a,  b,  c*  Then  we 
have  (Art.  14)  the  following  relations :  ^  I  'c 


a  =  csin  A  =  ccosB  =  6  tan  A  =  &cotB     .     .     (1) 

b  =csinB  =  ccosA=  atanB  =  a  cot  A     .     .     (2) 

c  =  6  sec  A  =  asecB  =  frcosecB  =  acosec  A  .     (3) 

which  may  be  expressed  in  the  following  general  theorems : 

*  The  student  must  remember  that  a,  6,  c,  are  numbers  expressing  the  lengths  of 
the  sides  in  terms  of  some  unit  of  length,  such  as  a  foot  or  a  mile.  The  unit  may  be 
whatever  we  please,  but  must  be  the  same  for  all  the  sides. 


OBLIQUE   TRIANGLES. 


147 


I.  In  a  right  triangle  each  side  is  equal  to  the  product  of  the 
hypotenuse  into  the  sine  of  the  opposite  angle  or  the  cosine  of 
the  adjacent  angle. 

II.  In  a  right  triangle  each  side  is  equal  to  the  product  of 
the  other  side  into  the  tangent  of  the  angle  adjacent  to  that 
other  side,  or  the  cotangent  of  the  angle  adjacent  to  itself. 

III.  In  a  right  triangle  the  hypotenuse    is   equal  to   the 
product  of  a  side  into  the  secant  of  its  adjacent  angle,  or  the 
cosecant  of  its  opposite  angle. 

EXAMPLES. 

In  a  right  triangle  ABC,  in  which  C  is  a  right  angle, 
prove  the  following : 

1.   tan  B  =  cot  A  +  cos  C.  2.   sin2A  =  sin2B. 

3.   cos2A  +  cos2B  =  0.  4. 


7.   tan2A  = 


a        b 
~2b     2a 
2ab 

& 

ft2          ~2 

6.   cos2A  =  —  --• 
c2 

8.    RinSA-3^'-"3, 

b2  -  a? 


c3 


OBLIQUE  TRIANGLES. 


95.  Law  of  Sines.  —  In  any  triangle  the   sides  are  pro- 
portional to  the  sines  of  the  opposite  angles. 

Let  ABC  be  any  triangle.  Draw 
CD  perpendicular  to  AB. 

We  have,  then,  in  both  figures 

CD  «=  a  sin  B  =  b  sin  A.     (Art.  94) 
.•.  asinB  =  b  sin  A. 

a  b 

sin  A      sin  B 

Similarly,  by  drawing  a  perpen- 
dicular from  A  or  B  to  the  opposite 
side,  we  may  prove  that 


148 


PLANE  TRIGONOMETRY. 


or 


sinB 
a 


sinC 
b 


and 


c 


sin  C      sin  A 


sin  A      sinB      sinC 
a  :  6  :  c  =  sin  A  :  sin  B  :  sin  C. 


96.  Law  of  Cosines.  —  In  any  triangle  the  square  of  any 
side  is  equal  to  the  sum  of  the  squares  of  the  other  two  sides 
minus  twice  the  product  of  these  sides  and  the  cosine  of  the 
included  angle. 

In  an  acute-angled  triangle  (see 
first  figure)  we  have  (Geom.,  Book 
III.,  Prop.  26) 

BC2  =  AC2  +  AB2-2AB  x  AD, 

A' 


or     a  = 


But 


AD  =  b  cos  A. 


In  an  obtuse-angled  triangle  (see 
second  figure)  we  have  (Geom., 
Book  III.,  Prop.  27) 


D  A 

BC2  =  AC2  +  AB2  +  2ABx  AD, 


c          B 


or 


AD  =  b  cos  CAD  =  -  b  cos  A. 


But 


Similarly,       b2  =  c2  +  a2  -  2  ca  cos  B, 


NOTE.  —  When  one  equation  in  the  solution  of  triangles  has  been  obtained,  the 
other  two  may  generally  be  obtained  by  advancing  the  letters  so  that  a  becomes  b, 
b  becomes  c,  and  c  becomes  a;  the  order  is  abc,  bca,  cab.  It  is  obvious  that  the 
formulae  thus  obtained  are  true,  since  the  naming  of  the  sides  makes  no  difference, 
provided  the  right  order  is  maintained. 


OBLIQUE  TRIANGLES.  149 

97.  Law  of  Tangents.  —  In  any  triangle  the  sum  of  any 
two  sides  is  to  their  difference  as  the  tangent  of  half  the  sum 
of  the  opposite  angles  is  to  the  tangent  of  half  their  difference. 

By  Art.  95,     a  :  b  =  sin  A  :  sin  B, 
By  composition  and  division, 

a  +  b  __  sin  A  -f-  sin  B 
a  —  b      sin  A  —  sin  B 


c-a     tan(C-A) 

Since  tan£(A  +  B)  =  tan  (90°  -  £C)  =  co 
the  result  in  (1)  may  be  written 

a  +  b  _        cot^C  ,£. 

a-b      tanJ(A-B) 

and  similar  expressions  for  (2)  and  (3). 

98.   To  show  that  in  any  triangle  c  =  a  cos  B  +  b  cos  A. 

In  an  acute-angled  triangle  (first  figure  of   Art.  96)   we 
have 


=  a  cos  B  +  6  cos  A. 

In  an  obtuse-angled  triangle   (second  figure  of  Art.  96) 
we  have  C  =  DB-DA 

=  a  cos  B  —  b  cos  CAD. 
.*.  c  =  acosB  +  6  cos  A. 
Similarly,  6  =  c  cos  A  +  a  cos  C, 

a  =  b  cos  C  -f-  c  cos  B. 


150  PLANE  TRIGONOMETRY. 


EXAMPLES. 

1.  In  the  triangle  ABC  prove  (1) 

a  +  b:c  =  cos  ^  (A  —  B)  :  sin  1  C, 
and  (2)  a  —  b:c  =  sin  \  (  A  -  B)  :  cos  1  C. 

2.  If  AD  bisects  the  angle  A  of  the  triangle  ABC,  prove 

BD  :  DC  =  sin  C  :  sin  B. 

3.  If  AD'  bisects  the  external  vertical  angle  A,  prove 

BD':CD'  =  sinC:sinB. 

4.  Hence  prove    J-  ^cos^Acos  *  (B  -  C) 

DC  asinB 

and  also  JL  =  2sin|  Asini(C  -  B). 

D'C  asinB 

99.   To  express  the  Sine,  the  Cosine,  and  the  Tangent  of 
Half  an  Angle  of  a  Triangle  in  Terms  of  the  Sides. 

I.    By  Art.  96  we  have 

cos  A  =  b*  +  °2  ~  a"  '  =  1  -  2  sin2-  .     .     •     •     (Art.  49) 

Zi  OC  £ 


2bc 


2  be 

_  (a  +  b  —  c)  (a  —  b  +  c) 

2  be 
Let  a-|-Z>-|-c  =  2s; 

then  a  +  b  —  c  =  2(s—  c),  and  a  —  6  +  c  =  2(s  —  b). 


2  2bc 


OBLIQUE  TRIANGLES. 


151 


Similarly,     sm^  =  A/-v-  — '- (2) 

Ln2 

II.  cosA  =  2cos2--! (Art.  49) 

S  2  ~  2  be 

2  be 
_  (a  +  b  -f-  e)  (b  -f-  c  —  a)       * 

2  be 
=  2s'2(s-a) 

2  be 

...  cos|=     ^7 
Similarly,    Cos?  =  A/^--^.     .          .....     (5) 

III.  Dividing  (1)  by  (4),  we  get 

r        /77 — ^TT; — ^\ 

J_  v  /  \  "  "~~"  Cv  I  I  O   "" "• "   C/  I  /  f\\ 

2  =  V /   _   x 

Since  any  angle  of  a  triangle  is  <  180°,  the  half  angle  is 
<90°;  therefore  the  positive  sign  must  be  given  to  the 
radicals  which  occur  in  this  article. 


152  PLANE  TRIGONOMETRY. 

100.  To  express  the  Sine  of  an  Angle  in  Terms  of  the 
Sides. 

sin  A  =  2  sin  A  cos         .....     (Art.  49) 

_ 


2     ,  ___ 
.-.  sin  A  =  —  V  s  (s  —  a)  (s  —  6)  (s  —  c). 
be 

2      .  ____  . 
Similarly,   sin  B  =  —  Vs  (s  —  a)  (s  —  b)  (s  —  c), 


ac 
o 


ab 

Cor.  sin  A  ==  —  V2  b*c?+~2cW+  2  a262-  a4-¥^c"4, 

and  similar  expressions  for  sin  B,  sin  C. 

EXAMPLES. 

In  any  triangle  ABC  prove  the  following  statements : 
1.    a  (6  cos  C  —  ccosB)  =  62  —  c2. 

o     sin  A  -f-  2  sin  B  _  sin  C 

sin2  A  —  m  sin2  B      sin2  C 

4.  — — —  = • 

a2  —  ra&2  c2 

5.  a  cos  A  -f&cosB  —  ccosC  =  2ccos  AcosB. 
K         cos  A       •       cosB        cos  C  0 


sin  B  sin  C      sin  C  sin  A      sin  A  sin  B 

7.  a  sin  (B  -  C)  +  6  sin  (C  -  A)  +  c  sin  ( A  -  B)  =  0. 

8.  tan  £  A  tan  £  B  =  ^^. 

s 

9.  taniA-s-taniB=.(s-&)-s-(*-c). 


AEEA   OF  A   TRIANGLE. 


153 


101.  Expressions  for  the  Area 
of  a  Triangle. 

(1)  Given  two  sides  and  their 
included  angle. 

Let  S  denote  the  area  of  the  tri- 
angle ABC.  Then  by  Geometry, 

2S  =  cxCD. 


But   in   either   figure,   by   Art. 


94, 


=  6sinA. 


D 


Similarly,         S  =  ^  ac  sin  B, 
S  =  i  ab  sin  C. 

(2)  Given  one  side  and  the  angles. 
Since 


which  is 


Similarly, 


a  :  b  =  sin  A  :  sin  B (Art.  95) 

sin  A  ' 
S  =  i  ab  sin  C,  gives 

q2sinB  sinG 
2  sin  A 

62  sin  A  sin  C      c2  sin  A  sin  B 


2  sin  B 
(3)   Given  the  three  sides. 


2sinC 


sin  A  =  ~  Vs(s  -  a)  (s  -  b)  (s  -  c)    (Art.  100) 
be 


Substituting  in 


S  =  \  be  sin  A, 


we  get 


S  =  Vs(s  -  a)  (s  -  b}  (s  -  c) . 


154 


PLANE  TRIGONOMETRY. 


102.   Inscribed  Circle.  —  To  find  the  radius  of  the  inscribed 
circle  of  a  triangle. 

Let  ABC  be  a  triangle,  0  the 
centre  of  the  inscribed  circle,  and 
r  its  radius.  Draw  radii  to  the 
points  of  contact  D,  E,  F ;  and  join 
OA,  OB,  OC.  Then 


S  =  area  of  ABC 
=  A  AOB  +  A  BOC  +  A  COA 


=  r 


=  rs 


(Art.  99) 


103.   Circumscribed  Circle.  —  To  find  the  radius  of  the 
circumscribed   circle  of  a   triangle   in  ^^__^« 

terms  of  the  sides  of  the  triangle. 

Let  0  be  the  centre  of  the  circle    A/ 
described   about   the   triangle    ABC, 
and  R  its  radius. 

Through  0  draw  the  diameter  CD 
and  join  BD. 

Then        Z  BDC  =  Z  BAG  =  Z  A. 


=  2KsinA,  or  a  =  2RsinA. 

=      a      =      b      =      c 
2  sin  A      2  sin  B      2  sin  C 


But 


be 


(1) 

(Art.  101) 
(2) 


RADII  OF  THE  ESCRIBED    CIRCLES. 


155 


104.  Escribed  Circle.  —  To  find 
the  radii  of  the  escribed  circles  of  a 
triangle. 

A  circle,  which  touches  one  side 
of  a  triangle  and  the  other  two 
sides  produced,  is  called  an  escribed 
circle  of  the  triangle. 

Let  0  be  the  centre  of  the 
escribed  circle  which  touches  the 
side  BC  and  the  other  sides  pro- 
duced, at  the  points  D,  E,  and  F, 
respectively,  and  let  the  radius  of 
this  circle  be  ?v 

We  then  have  from  the  figure 

A  ABC  =  A  AOB  +  A  AOC  -  A  BOG. 
Q  _  cri  i  bri     ari 


S 


s  —  a 


+  c  -  a)  =  r,  (s  -  a)  .     (Art.  99) 
.     .     (1) 


Similarly  it  may  be  proved  that  if  r2,  rs  are  the  radii  of 
the  circles  touching  AC  and  AB  respectively, 

S  S 


—  c 


105.  To  find  the  Distance  be- 
tween the  Centres  of  the  Inscribed 
and  Circumscribed  Circles*  of  a 
Triangle. 

Let  I  and  0  be  the  incentre 
and  circumcentre,  respectively,  of 
the  triangle  ABC,  IA  and  1C 
bisect  the  angles  BAG  and  BC  A  ; 

*  Often  called  the  incentre  and  circumcen- 
tre of  a  triangle. 


156  PLANE  TRIGONOMETRY. 

therefore  the  arc  BD  is  equal  to  the  arc  DC,  and  DOH 
bisects  BC  at  right  angles. 

Draw  IM  perpendicular  to  AC.     Then 

Z  DIC  =  A-±L^  =  BCD  +  BCI  =  DCI. 


Also,  AI  =  IMcosec—  =  rcosec—  • 

2  2 


that  is,  (K  +  01)  (E  -  01)  =  2  Kr. 


EXAMPLES. 


1.  The  sides  of  a  triangle  are  18,  24,  30  ;  find  the  radii 
of  its  inscribed,  escribed,  and  circumscribed  circles. 

Ans.  6,  12,  18,  36,  15. 

2.  Prove  that  the  area  of  the  triangle  ABC  is 


2  cot  A  +  cot  B 

3.   Find  the  area  of  the  triangle  ABC  when 

(1)  a  =  4,  b  =  10  ft.,  C  =  30°.  Ans.  10  sq.  ft. 

(2)  6=5,  c  =  20  inches,  A  =  60°.  43.3  sq.  in. 

(3)  a  =  13,  b  =  14,  c  =  15  chains.         84  sq.  chains. 


4.  Prove     =     +     +    . 

r  r      r 


5.  Prove  r  =  -_ 

cosiA 


EXAMPLES.  157 

6.  Prove  that  the  area  of  the  triangle  ABC  is  represented 
by  each  of  the  three  expressions  : 

2  R2  sin  A  sin  B  sin  C, 

rs,  and 
Rr(siii  A  +  sin  B  +  sin  C). 

7.  If  A  =  60°,  a  =  V3,  6=V2,  prove  that  the  area 

=  i(3+V3). 

8.  Prove  R  (sin  A  +  sin  B  -+-  sin  C)  =  s. 

9.  Prove  that  the  bisectors  of  the  angles  A,  B,  C,  of  a 
triangle  are,  respectively,  equal  to 

2  be  cos  —    2  ca  cos  -    2  ab  cos  - 

222 


— ,    •  — ,   - 

o-fc  c-\-  a  a-{- 

106.  To  find  the  Area  of  a  Cyclic* 
Quadrilateral.  A 

Let  ABCD  be  the  quadrilateral,  and 
a,  b,  c,  and  d  its  sides.     Join  BD. 
Then,  area  of  figure  =  S 

=  ^  ad  sin  A  -f  ^  be  sin  C 
=  ^(ad  +  6c)sin  A     .     .     .     (1) 
Now  in  A  ABD,    BD2  =  a2  +  d2  -  2  ad  cos  A, 
and  in  A  CBD,          51?=  62  +  c2  -  26ccos  C 
=  62  +  c2  —  2  be  cos  A. 


.  ra2_&2_c2_h^ 

'•SmA=V1-         2(ad  +  5c) 


-  (a2  -  52  -  c2 


*  See  Geometry,  Art.  251. 


158  PLANE  TRIGONOMETRY. 


2  (ad  -f  be) 


=  2V  (s  -  a)  (s  -  6)  (8  -  c)  (s  -  d) 
ad  -f  &c 


(where  2s  =  a-|-&  + 
Substituting  in  (1),  we  have 


S  =  V  (s  -  a)  (s  -  b)  (s  -c)(s-d). 

The  more  important  formulae  proved  in  this  chapter  are 
summed  up  as  follows  : 


1.    _^_  =  -_  =  _^_  (Art.  95) 

sin  A      sin  B      sin  C 

•2.    a2=&2  +  c2-26ccosA  .......     (Art.  96) 


a==  (Art.  97) 

a-  5      tan^(A-B) 


4.    BiniA=:J<'-ft)<'-c>.     .....     (Art.  99) 

\  be 


Is  (s  —  a) 

5. 


6. 


7.    sin  A    =-^s(s-a)(s-b)(s-c)    .     .   (Art.  100) 
oc 


—  V2  62c2  +  2  c2a2  -f-  2  a262  -  a4  -  64  -  c4. 
8.    Area  of  A  =  Vs  (s  -  a)  (s-b)  (s-c).     .   (Art.  101) 


EXAMPLES.  159 

9.    Area  of  A  =  £  (a  +  6  +  c)  =  rs     .     .     .     .   (Art.  102) 


10. 

S 


—  a)  (s  —  b)  (s  —  c) 
s 

11.  K  =  — (Art.  103) 


EXAMPLES. 


In  a  right  triangle  ABC,  in  which  C  is  the  right  angle, 
prove  the  following  : 


0 

2- 


sin2  A  +  sin2B 
.  9B     c  —  a 


3.      cos 


9A      b  4-  c 

4.  cos2—  =  — - — 

2        2c 

5.  sin(A-B)  +  cos2A  = 

-.  .       1.  A  T> 

6. 


2 
7.    sin(A-B)  +  sin(2A 

8. 


c 
9.    (sin  A  -  sin  B)2+  (cos  A  +  cos  B)2  =  2. 

^  Q        /a  •  +  &  i      lQ>  —  b_    2  sin  A 
\a  —  6      \a  +  6      Vcos2B 

In  any  triangle  ABC,  prove  the  following  statements  : 
11. 


160  PLANE  TRIGONOMETRY. 


12.    (&  -c)  COT  —  =  a  sin 


2  2 

a  —  b      cos  B  —  cos  A 


14. 


c  1  4  cos  C 


1  ~     b  4  c     cos  B  -f-  cos  C 
lo.    -  —  —  -  -  -  —  • 
a  1  —  cos  A 

16. 


b  +  c 

17.  a-f-6+c  =  (6  4  c)cos  A  4  (c  4  a)cos  B  4-  (a  4  6)cos  C. 

18.  b  +  c—  a  =  (6  +  c)cos  A  —  (c  —  a)cosB  -f  (a  —  6)cos  C. 

19.  a  cos  (A  +  B  -f  C)  -  b  cos  (B  +  A)  -  c  cos  (  A  +  C)  =  0. 
orv    cos  A      cos  B      cos  C  _  a2  -f-  62  4-  c2 


a  b  c  2abc 

A          a  sin  C 
21.   tanA  =  -  — -• 

b  —  a  cos  C 

n  TI 

22. 


2  2 

T?        n       ^  \ 
23. 


2        2      b+c+a 

24.  tan-(&4c-a)  =  tan| 

25.  c2«=(a  +  6)2sin22  +  (a-6)2cos2-. 

2  2 

26.  ccosA4-cosB  =  2a4-&sin25. 


27. 

28.   tan  B  -*-  tan  C  =  (a2  +  b2  -  c2)  -*-  (a2  -  b2  4  c2). 


EXAMPLES.  161 

29.  a2  -f  b'2  +  c2  =  2(a6  cos  C  +  6c  cos  A  +  ca  cos  B). 

30.  cos2-^-cos2?  =  (s-a)--6(s-6). 

2  2 

r^  Ti 

31.  6  sin2  —  |-  c  sin2  —  =  s  —  a. 

2  Z 

32.  If  p  is  the  length  of  the  perpendicular  from  A  on 

BC,  sinA  =  ^. 

be 

33.  If  A  =  3B, 


6'gin  B 


34.  If  V6c  sin  B  sinC  =  ,  then  B  =  C. 

b  +  c 

B       C          A 

35.  a  cos—  cos  —  cosec—  =  s. 

36.  If  cos  A  =  |,  and  cos  B  =  |f  ,  then  cos  C  =  —  if. 

37.  If  sin2B  -f  sin2C  =  sin2  A,  then  A  =  90°. 

38.  If  D  is  the  middle  point  of  BC,  prove  that 


39.  If  a  =  26,  and  A  =  3B,  prove  that  C  =  60°. 

40.  If  D,  E,  F,  are  the  middle  points  of  the  sides,  BC,  CA, 
AB,  prove 

4(AIT  +  BE2  +  CF2)  =  3(a2  +  62  +  c2). 

41.  If  a,  6,  c,  the  sides  of  a  triangle,  are  in  arithmetic 
progression,  prove 

tan  -  tan  5  =  !. 

2        2      3 

Af>    T-P  tan  A  —  tan  B      c  —  6  ,  ,  /.AO 

42.  If  -  —  =  --    prove  that  A  =  60  . 

tan  A  4-  tan  B    c  ' 

43.  If  cos  B  =  ^BA,  prove  that  B  =  C. 

2sinC    J 


162  PLANE  TRIGONOMETRY. 

44.  If  a2  =  b2  -  be  +  c2,  prove  that  A  =  60°. 

45.  If  the  sides  of  a  triangle  are  a,  6,  and  V«2  +  ab  -f  62, 
prove  that  its  greatest  angle  is  120°. 

46.  Prove   that   the   vertical   angle   of   any   triangle   is 
divided  by  the  median  which  bisects  the  base,  into  seg- 
ments whose  sines  are  inversely  proportional  to  the  adja- 
cent sides. 

47.  If  AD  be  the  median  that  bisects  BC,  prove  (1) 

(b2  -  c2)  tan  ADB  =  2  be  sin  A, 
and  (2)        cot  BAD  +  cot  DAC  =  4  cot  A  +  cot  B  +  cot  C. 

48.  Find  the  area  of  the  triangle  ABC  when  a  =  625, 
b  =  505,  c  =  904  yards.  Ans.  151872  sq.  yards. 

49.  Find   the  radii  of   the   inscribed   and   each   of   the 
escribed  circles  of  the  triangle  ABC  when  a  =  13,  6  =  14, 
c  =  15.  Ana.  4;  10.5;  12;  14. 

50.  Prove  the  area  S  =  -|  a2  sin  B  sin  C  cosec  A. 

51.  "       "      "     "=Vnyvv 

52.  «        «       «     "=     2abc     /cos  -cos-  cos 


2        2 

53.  Prove  that  the  lengths  of  the  sides  of  the  pedal  tri- 
angle, that  is,  the  triangle  formed  by  joining  the  feet  of  the 
perpendiculars,  are  a  cos  A,  b  cos  B,  c  cos  C,  respectively. 

54.  Prove   that   the   angles  of   the   pedal   triangle   are, 
respectively,  TT  —  2  A,  IT  —  2  B,  TT  —  2  C. 

55.  Prove  r.r^  =  r8  cot2  -  cot2  -  cot2  -. 

A  T?      r1 

56.  Prove  r±  cos  —  =  a  cos  —  cos  — 

2  22 

57.  Prove  that  the  area  of  the  incircle  :  area  of  the  tri- 

»  T>  /~1 

angle  :  :  TT  :  cot  —  cot  —  cot  —  • 
222 


EXAMPLES.  163 

Prove  the  following  statements  : 

58.  If  a,  b,  c,  are  in  A.  P.,  then  ac  =  6  rR. 

59.  If  the  altitude  of  an  isosceles  triangle  is  equal  to  the 
base,  R  is  five-eighths  of  the  base. 

60.  be  =  4  R2(cos  A  +  cos  B  cos  C) . 

61.  If  C  is  a  right  angle,  2  r  +  2  R  =  a  +  5. 

62.  r2r3  +  ?v*i  +  Wz  =  s2- 

63.  1  +  1  +  1= 


be     ca     ab     2rR 

C 

64.  i1!  -f-  r2  =  ccot— • 

A  B   .    C 

65.  r  cos  —  =  a  sin  —  sin  — 

2  22 

66.  If  pu  pz,  p3  be  the  distances  to  the  sides  from  the 
circumcentre,  then 

a_.   b^  ,  _c  __abc_ 

Pi     Pi     Ps     ^PiPiPs 

67.  The  radius  R  of  the  circumcircle 


1  s/  _  abc  _ 

2  Af  sin  A  sin  B  sinC 


68.    S  =  -sin2B-f-sin2A. 
4  4 

69. 


s  —  a     s  —  6      s  —  c      s        S 

70.  a6cr  =  4R(s-a)(s-6)(s-c). 

71.  The  distances  between  the  centres  of  the  inscribed 

j^ 
and   escribed   circles  of  the   triangle   ABC  are   4  R  sin  —  , 

4Rsin?,  4Rsin^. 

2'  2 

72.  If  A  is  a  right  angle,  r2  +  r3  =  a. 


164  PLANE  TRIGONOMETRY. 

73.  In  an  equilateral  triangle  3  K  =  6  r  =  2  ?*j. 

74.  If  r,  rly  r2,  rs  denote  the  radii  of  the  inscribed  and 
escribed  circles  of  a  triangle, 


- 

2      T2r3 

75.  The  sides  of  a  triangle  are  in  arithmetic  progression, 
and  its  area  is  to  that  of  an  equilateral  triangle  of  the  same 
perimeter  as  3  is  to  5.     Find  the  ratio  of  the  sides  and  the 
value  of  the  largest  angle.  Ans.  As  7,  5,  3  ;  120°. 

76.  If    an    equilateral    triangle   be   described   with    its 
angular   points    on   the   sides    of    a   given   right    isosceles 
triangle,  and  one  side  parallel  to  the  hypotenuse,  its  area 
will  be  2  a2  sin2  15°  sin  60°,  where  a  is  a  side  of  the  given 
triangle. 

77.  If  h  be  the  difference  between  the  sides  containing 
the  right  angle  of  a  right  triangle,  and  S  its  area,  the  diam- 
eter of  the  circumscribing  circle  =  V^2  +  4S. 

78.  Three  circles  touch  one  another   externally  :    prove 
that  the  square  of  the  area  of  the  triangle  formed  by  join- 
ing their  centres  is  equal  to  the  product  of  the  sum  and 
product  of  their  radii. 

79.  On  the  sides  of  any  triangle  equilateral  triangles  are 
described   externally,  and  their  centres  are  joined:   prove 
that  the  triangle  thus  formed  is  equilateral. 

80.  If  G!,  02,  03  are  the  centres  of  the  escribed  circles  of 
a  triangle,  then  the  area  of  the  triangle  OiOsOg  =  area  of 

triangle  ABcfl  +  j-y-5  -  +         b         +  --  c-  -  1- 

6  -f-  c  —  a      a  +  c  —  b      a  +  b  —  c  J 

81.  If  the  centres  of  the  three  escribed  circles  of  a  tri- 
angle are  joined,  then  the  area  of  the  triangle  thus  formed 

is  —  -  where  r  is  the  radius  of  the  inscribed  circle  of  the 
,2r 

original  triangle. 


SOLUTION  OF  TRIANGLES.  165 


CHAPTER  VII. 

SOLUTION  OP  TKIATOLES, 

107.  Triangles,  —  In  every  triangle  there  are  six  elements, 
the  three  sides  and  the  three  angles.     When  any  three  ele- 
ments are  given,  one  at  least  of  the  three  being  a  side,  the 
other  three  can  be  calculated.     The  process  of  determining 
the  unknown  elements  from  the  known  is  called  the  solution 
of  triangles. 

NOTE.  — If  the  three  angles  only  of  a  triangle  are  given,  it  is  impossible  to  deter- 
mine the  sides,  for  there  is  an  infinite  number  of  triangles  that  are  equiangular  to 
one  another. 

Triangles  are  divided  in  Trigonometry  into  right  and 
oblique.  We  shall  commence  with  right  triangles,  and  shall 
suppose  C  the  right  angle. 

RIGHT  TRIANGLES. 

108.  There  are  Four  Cases  of  Right  Triangles. 

I.  Given  one  side  and  the  hypotenuse. 

II.  Given  an  acute  angle  and  the  hypotenuse. 

III.  Given  one  side  and  an  acute  angle. 

IV.  Given  the  two  sides. 

Let  ABC  be  a  triangle,  right-angled 
at  C,  and  let  a,  &,  and  c,  as  before,  be 
the  sides  opposite  the  angles  A,  B, 
and  C,  respectively.  A  b 

The  formulae  for  the  solution  of  right  triangles  are  (1), 
(2),  (3)  of  Art.  94. 


166  PLANE   TRIGONOMETRY. 

109.   Case  I.  —  Given  a  side  and  the  hypotenuse,  as  a  and 
c;  to  find  A,  B,  b. 

We  have  sin  A  =  -. 

c 

.  *.  log  sin  A  =  log  a  —  log  c, 

from  which  A  is  determined  ;  then  B  =  90°  —  A. 
Lastly,  b  =  c  cos  A. 

.-.  log  b  =  log  c  -f-  log  cos  A. 
Thus  A,  B,  and  b  are  determined. 
Ex.  1.    Given  a  =  536,  c  =  941 ;  find  A,  B,  b. 

Solution  by  Natural  Functions. 

We  have         sin  A  =  -  =  —  =  .569607. 
c      941 

From  a  table  of  natural  sines  we  find  that 

A  =  34°  43' 22".     .-.  B  =  55°  16' 38". 

• 

Lastly,  b  =  c  cos  A  =  941  x  .821918 

=  773.425. 

Logarithmic  Solution. 


log  sin  A  =  log  a  —  log  c. 

log  o  =  2.7291648 
log  c  =  2.9735 


log  sin  A*  =  9.7555752 
.-.  A  =  34°  43' 22". 
.-.  B  =  55°  16' 38". 


log  b  =  log  c  +  log  cos  A. 

log  c  =  2.9735896 
log  cos  A  =  9.9148283 

log  b  =  2.8884179 
.'•.  6  =  773.424. 


Our  two  methods  of  calculation  give  results  which  do  not 
quite  agree.  The  discrepancies  arise  from  the  defects  of 
the  tables. 

*  Ten  is  added  so  as  to  get  the  tabular  logarithms  (Art.  76). 


EIGHT  TRIANGLES.  167 

The  process  of  solution  by  natural  sines,  cosines,  etc.,  can 
be  used  to  advantage  only  in  cases  in  which  the  measures 
of  the  sides  are  small  numbers. 

We  might  have  determined  b  thus : 


b  =  V(c-o)(c  +  o); 
or  thus  :  b  =  a  tan  B. 

NOTB.  —  It  is  generally  better  to  compute  all  the  required  parts  from  the  given 
ones,  so  that  if  an  error  is  made  in  determining  one  part,  that  error  will  not  affect  the 
computation  of  the  other  parts. 

To  test  the  accuracy  of  the  work,  compute  the  same  parts  by  different  formulae. 

Ex.  2.    Given  a  =  21,  c  =  29  ;  find  A,  B,  b. 

Ans.  A  =  46°  23'  50",  B  =  43°  36'  10",  b  =  20. 

NOTE.  — In  these  examples  the  student  must  find  the  necessary  logarithms  from 
the  tables. 

110.  Case  II.  —  Given  an  acute  angle  and  the  hypotenuse, 
as  A  and  c;  to  find  B,  a,  b, 

We  have  B  =  90°  -  A. 

Also  a  =  c  sin  A,  and  b  =  c  cos  A. 

.-.  log  a  =  log  c  +  log  sin  A  ; 
and  log  6  =  log  c  H-  log  cos  A. 

Thus  B,  a,  and  b  are  determined. 

Ex.  1.    Given       A  =  54°  28',  c  =  125 ;  find  B,  a,  b. 
B  =  90°  -  A  =  35°  32'. 

Solution  by  Natural  Functions. 

We  have  a  =  c  sin  A,  and  b  =  c  cos  A. 

Using  a  table  of  natural  sines,  we  have 

a  =  125  x  .813778  =  101.722, 
and  b  =  125  x  .581177  =    72.647. 


168 


PLANE   TRIGONOMETRY. 


Logarith  m ic  Solut ion. 


log  a  =  log  c  -f-  log  sin  A. 

log  c  =  2.0969100 
log  sin  A  =  9.9105057 

log  a*  =  2.0074157 
.-.  a  =  101. 722. 


log  b  =  log  c  +  log  cos  A. 

log  c  =  2.0969100 
log  cos  A  =  9.7643080 

log  b  *  =  1.8612180 
.-.  6  =  72.647. 


Ex.  2.    Giveji  A  =  37°  10',  c  =  8762 ;  find  a  and  b. 

Ans.  5293.4;  6982.3. 

111.   Case  III.  —  Given  a  side  and  .an  acute  angle,  as  A 
and  a;  to  find  B,  6,  c. 

We  have          B  =  90°  -  A. 


Also 


-,  and  c  =  — 


and 


tan  A  sin  A 

.-.  log  b  =  log  a  —  log  tan  A, 

log  c  =  log  a  —  log  sin  A. 


Ex.  1.   Given  A  =  32°  15'  24",  a  =  5472.5 ;  find  B,  b,  c. 

Solution. 
B  =  90°  -  A  =  57°  44'  36". 


log  b  =  log  a  —  log  tan  A. 

log  a  =  3.7381858 
log  tan  A  =  9.8001090 

log  b  =  3.9380768 
.-.  5  =  8671.152. 


log  c  =  log  a  —  log  sin  A. 

log  a  =  3.7381858 
log  sin  A  =  9.7273076 

log  c  =  4.0108782 
.«.  c=  10253.64. 


Ex.  2.    Given  A  =  34°  18',  a  =  237.6 ;  find  B,  6,  c. 

Ans.  B  =  55°42';  6  =  348.31;  c  =  421.63. 


*  Ten  ia  rejected  because  the  tabular  logarithmic  functions  are  too  large  by  ten 
(Art  76). 


EIGHT  TRIANGLES.  169 

112.   Case  IV. —  Given  the  two  sides,  as  a  and  b;  to  find 
A,  B,  c. 

We  have  tan  A  =  - ;  then  B  =  90°  -  A. 

b 

Also  c  =  a  cosec  A  =  — - — 

sin  A 

.*.  log  tan  A  =  log  a  —  log  6, 
and  log  c  =  log  a  —  log  sin  A. 

Ex.    Given  a  =  2266.35,  b  =  5439.24 ;  find  A,  B,  c. 
Solution. 


log  tan  A  =  log  a  —  log  b. 

log  a  =  3.3553270 
log  b  =  3.7355382 

log  tan  A  =  9.6197888 
.-.  A  =  22°  37' 12". 
.-.  B  =  67°  22' 48". 


log  c  =  log  a  —  log  sin  A. 

log  a  =  3.3553270 
log  sin  A  =  9. 5850266 

log  c  =  3.7703004 
.-.  c  =  5892.51. 


NOTE.  —  In  this  example  we  might  have  found  c  by  means  of  the  formula 
c  =  \/a2  +  62;  but  we  would  have  had  to  go  through  the  process  of  squaring  the 
values  of  a  and  b.  If  these  values  are  simple  numbers,  it  is  often  easier  to  find  c  in 
this  way;  but  this  value  of  c  is  not  adapted  to  logarithms.  A.  formula  which  consists 
entirely  of  factors  is  always  preferred  to  one  which  consists  of  terms,  when  any  of 
those  terms  contain  any  power  of  the  quantities  involved. 

113.  When  a  Side  and  the  Hypotenuse  are  nearly  Equal. 

—  When  a  side  and  the  hypotenuse  are  given,  as  a  and  c 
in  Case  I.,  and  are  nearly  equal  in  value,  the  angle  A  is  very 
near  90°,  and  cannot  be  determined  with  much  accuracy 
from  the  tables,  because  the  sines  of  angles  near  90°  differ 
very  little  from  one  another  (Art.  81).  It  is  therefore 
desirable,  in  this  case,  to  find  B  first,  by  either  of  the 
following  formulae : 


°fr <" 


1TO  PLANE  TRIGONOMETRY. 

rJ5°l? (Art.  50) 

1  +  cosB 

^~a  (2) 


-v 


Then  6  =  ccosA (3) 


or  =  V(cH-a)(c  —  a) (4) 

Ex.1.    Given  a  =  4602.21059,  c  =  4602.836;  find  B. 

c  -  a  =  0.62541,  log  (c  -  a)  =  T.7961648 

2  c  =  9205.672,  log  2  c  =  3.9640555 

2)5.8321093 

log  sin  ?  =  7.9160547 
& 

B  =  56'40".36.  .-.   ?  =  28'20".18. 

NOTE.  — The  characteristic  5  is  increased  numerically  to  6  to  make  it  divisible 
by  2  (see  Note  4  of  Art.  66).  Ten  is  then  added  to  the  characteristic  3,  making  it  7, 
so  as  to  agree  with  the  Tables  (Art.  76). 

There  is  a  slight  error  in  the  abore  value  of  B  on  account 
of  the  irregular  differences  of  the  log  sines  for  angles  near 
0°  (Art.  81).  A  more  accurate  value  may  be  found  by  the 
principle  that  the  sines  of  small  angles  are  approximately 
proportional  to  the  angles  (Art.  130). 

EXAMPLES. 

The  following  right  triangles  must  be  solved  by  log- 
arithms. 

1.  Given  a  =  60,  c  =  100 ;  find  A,  B,  b. 

Ans.  A  =  36°  52';  B  =  53°8';  6  =  80. 

2.  Given  a  =  137.66,  c  =  240  ;  find  A,  B,  b. 

Ans.  A  =  ,35° ;  B  =  55° ;  b  =,  196.59. 


EIGHT  TRIANGLES.  171 

3.  Given  a  =  147,  c  =  184 ;  find  A,  B,  b. 

Ana.  A  =  53°1'35";  B  =  36°  58' 25";  6  =  110.67. 

4.  Given  a  =  100,  c  =  200 ;  find  A,  B,  6. 

Ans.  A  =  30° ;  B  =  60° ;  b  =  100  V3. 

5.  Given  A  =  40°,  c  =  100  ;  find  B,  a,  6. 

Ans.  B  =  50°;  a  =  64.279;  6  =  76.604. 

6.  Given  A  =  30°,  c  =  150 ;  find  B,  a,  b. 

Ans.  B  =  60°;  a  =  75;  b  =  75 V3. 

7.  Given  A  =  32°,  c  =  1760;  find  B,  a,  b. 

Ans.  B  =  58°  ;  a  =  932.66  ;  b  =  1492.57. 

8.  Given  A  =  35°  16'  25",  c  =  672.3412 ;  find  B,  a,  b. 

Ans.  B  =  54°  43' 35";  a  =  388.26;  6  =  548.9. 

9.  Given  A  =  75°,  a  =  80;  find  B,  6,  c. 

Ans.  B  =  15°;  6  =  80(2-V3);  c  =  80(V6-V2). 

10.  Given  A  =  36°,  a  =  520 ;  find  B,  6,  c. 

Ans.  B  =  54°;  6  =  715.72;  c  =  884.68. 

11.  Given  A  =  34°  15',  a  =  843.2 ;  find  B,  6,  c. 

Ans.  B  =  55°45';  c  =  1498.2. 

12.  Given  A  =  67°  37'  15",  6  =  254.73 ;  find  B,  a,  c. 

Ans.  B  =  22°  22' 45";  a  =618.66;  c  =  669.05. 

13.  Given  a  =  75,  6  =  75 ;  find  A,  B,  c. 

Ans.  A  =  45°  =  B ;  c^=  75  V2. 

14.  Given  a  =  21,  6  =  20 ;  find  A,  B,  c. 

Ans.  A  =  46°  23' 50";  c  =  29. 

15.  Given  a  =  300.43,  6  =  500 ;  find  A,  B,  c. 

Ans.  A  =  31°;  B  =  59°;  c  =  583.31. 

16.  Given  a  =  4845,  6  =  4742;  find  A,  B,  c. 

Ans.  A  =  45°  36' 56". 


172  PLANE  TRIGONOMETRY. 

OBLIQUE  TRIANGLES. 

114.  There  are  Four  Cases  of  Oblique  Triangles. 

I.  Given  a  side  and  two  angles. 

II.  Given  two  sides  and  the  angle  opposite  one  of  them. 

III.  Given  two  sides  and  the  included  angle. 

IV.  Given  the  three  sides. 

The  formulae  for  the  solution  of  oblique  triangles  will  be 
taken  from  Chap.  VI.  Special  attention  must  be  given  to 
the  following  three,  proved  in  Arts.  95,  96,  97. 

a  b  c 


(1)  The  Sine-rule, 


sin  A      sin  B      sin  C 

?  +  c2-a2 


(2)  The  Cosine-rule,  cos  A  = 

2  be 

(3)  The  Tangent-rule,        tan^-^  =  ^=^cot-« 

2t          a  -\-  o       2i 

115.   Case  I.    Given  a  side  and  two  angles,  as  a,  B,  C;  find 
A,  b,  c. 

(1)         A  =  180°  -  (B  +  C).        .-.  A  is  found. 
b  a  ,       asinB 


sinB      sin  A  sin  A 

c  a  a  sin  C 


(2) 

(3) 

sinC      sin  A  sin  A 

These  determine  b  and  c. 

Ex.  1.    Given  a  =  7012.6,  B  =  38°  12' 48",  C=60°;  find 
A,  6,  c. 

Solution. 

A  =  180°  -  (B  +  C)  =  81°  47'  12". 


log  a  =  3.8458729 

log  sin  B  =  9.9714038 

colog*  sin  A  =  0.0044775 

log  b  =  &8217542 


b  =  6633.67. 


log  a  =  3.8458729 

log  sin  C  =  9.9375306 

colog  sin  A  =  0.0044775 

log  c  =  3.7878810 

.-.  c  =  6135.94. 


*  See  Art. 


OBLIQUE  TRIANGLES.  173 

Ex,  2.    Given  a=1000,  B  =  45°,  C  =  127°  19' ;  find  A,  b,  c. 
Ans.  A  =  7°  11';  6  =  5288.8;  c  =  5948.5. 

116.  Case  II.  Given  two  sides  and  the  angle  opposite  one 
of  them,  as  a,  b,  A;  find  B,  C,  c. 

(1)  sin  B  =  ftsmA;  thus  B  is  found. 

ft 

(2)  C  =  180°  _  (A  +  B);  thus  C  is  found. 

/o\  a.  sin  C     ,T_          •    f        •, 

(3)  c  =  —      — ;  thus  c  is  touna. 

sin  A 

This  is  usually  known  as  the  ambiguous  case,  as  shown  in 
geometry  (B.  II.,  Prop.  31).  The  ambiguity  is  found  in 
the  equation 


Since  the  angle  is  determined  by  its  sine,  it  admits  of 
two  values,  which  are  supplements  of  each  other  (Art.  29). 
Therefore,  either  value  of  B  may  be  taken,  unless  excluded 
by  the  conditions  of  the  problem. 

I.  If  a  <  6  sin  A,   sin  B  >  1,  which   is   impossible ;    and 
therefore  there  is  no  triangle  with  the  given  parts. 

II.  If  a  =  b  sin  A,  sin  B  =  1,  and  B  =  90° ;  therefore  there 
is  one  triangle  —  a  right  triangle  —  with  the  given  parts. 

III.  If  a  >  b  sin  A,  and  <  b,  sin  B  <  1 ;  hence  there  are 
two  values  of  B,  one  being  the  supplement  of  the  other, 
i.e.,  one  acute,  the  other  obtuse,  and  both  are  admissible ; 
therefore  there  are  two  triangles  with  the  given  parts. 

IV.  If  a  >  b,  then  A  >  B,  and  since  A  is  given,  B  must 
be  acute ;  thus  there  is  only  one  triangle  with  the  given 
parts. 


174 


PLANE   TRIGONOMETRY. 


These  four  cases  may  be  illustrated  geometrically. 

Draw  A,  the  given  angle.     Make  AC  =  b  ;  draw  the  per- 
pendicular CD,  which  =  b  sin 
A.    With  centre  C  and  radius 
a,  describe  a  circle. 

I.  If  a<6  sin  A,  the  circle 
will  not  meet  AX,  and  there- 
fore no  triangle  can  be  formed 
with  the  given  parts. 

II.  If  a  =  b  sin  A,  the  cir- 
cle touches  AX  in  B'  ;  there- 
fore   there   is   one   triangle, 
right-angled  at  B. 

III.  If    a  >  b  sin  A,    and 
<  b}  the  circle   cuts  AX  in 
two  points  B  and  B',  on  the  _ 
same  side  of  A;  thus  there  B/x     A 

are  two  triangles  ABC  and  AB'C,  each  having  the  given 
parts,  the  angles  ABC,  AB'C  being  supplementary. 

IV.  If  a  >  &,  the  circle  cuts  AX  on  opposite  sides  of  A, 
and  only  the  triangle  ABC  has  the  given  parts,  because  the 
angle  B'AC  of  the  triangle  AB'C  is  not  the  given  angle  A, 
but  its  supplement. 

These  results  may  be  stated  as  follows  : 


a  <  b  sin  A, 

a  *=  b  sin  A, 

a  >  b  sin  A  and  <  6, 

a  >  b, 


no  solution. 

one  solution  (right  triangle). 

two  solutions. 

one  solution. 


These  results  may  be  obtained  algebraically  thus  : 

We  have          a2  ==  W  +  c2  -  2  be  cos  A      .     .     .     (Art.  96) 


.•.  c  =  b  cos  A  ±  Va2  —  b2  sin2  A, 


OBLIQUE  TRIANGLES.  175 

giving   two   roots,    real  and   unequal,  equal  or  imaginary, 
according  as  a  >,  =,  or  <  b  sin  A. 

A  discussion  of  these  two  values  of  c  gives  the  same 
results  as  are  found  in  the  above  four  cases.  We  leave 
the  discussion  as  an  exercise  for  the  student. 

NOTE.  —  "When  two  sides  and  the  angle  opposite  the  greater  are  given,  there  can 
be  no  ambiguity,  for  the  angle  opposite  the  less  must  be  acute. 

When  the  given  angle  is  a  right  angle  or  obtuse,  the  other  two  angles  are  both 
acute,  and  there  can  be  no  ambiguity. 

In  the  solution  of  triangles  there  can  be  no  ambiguity,  except  when  an  angle  is 
determined  by  the  sine  or  cosecant,  and  in  no  case  whatever  when  the  triangle  has 
a  right  angle. 

Ex.  1.   Given  a  =  7,  6  =  8,  A  =  27°  47' 45";  find  B,  C,  c. 


Solution. 


log  b  =  0.9030900 

log  sin  A  =  9.6686860 

cologa  =  9.1549020 

log  sin  B  =  9.7266780 


log  a  =  0.8450980 

log  sin  C  =  9.9375306 

colog  sin  A  =  0.3313140 

log  c=  1.1139426 


.-.  B  =  32°  12' 15",  or  147°47'45".  .-.  c  =  13. 

.%  0  =  120°,  or  4°  24' 30". 
Taking  the  second  value  of  C  as  follows : 

log  a  =  0.8450980 

log  sin  C  =  8.8857232 

colog  sin  A  =  0.3313140 

log  c  =  0.0621352 
.-.  c  =  1.1538. 
Thus,  there  are  two  solutions.     See  Case  III. 

Ex.  2.    Given   a  =  31.239,   b  =  49.5053,   A  =  32°  18';   find 
,  C,  c.  Ans.  B  =  56°  56'  56".3,  or  123°    3'    3".7 ; 

C  =  90°45'    3".7,  or     24°38'56".3; 
c  =  58.456,  or  24.382. 


176  PLANE   TRIGONOMETRY. 

117.    Case  III.  —  Given  two  sides  and  the  included  angle, 
as  a,  b,  C;  find  A,  B,  c. 

(1)  tanB  =  cotC      .     .     (Art.114) 


Hence       =-     is  known,  and  =  90°  -    - 

2  22 

.\  A  and  B  are  found. 

c  =  asinC      r6sinC 
sin  A'          sinB' 

and  thus  c  is  found  and  the  triangle  solved. 

In  simple  cases  the  third  side  c  may  be  found  directly  by 
the  formula 

c  =  Va2  +  b2  -  2  ab  cos  C   .     .     .     (Art.  96) 

or  the  formula  may  be  adapted  to  logarithmic  calculation 
by  the  use  of  a  subsidiary  angle  (Art.  90). 

Ex.  1.    Given   a  =  234.7,   b  =  185.4,  C  =  84°  36'  ;   find  A, 
B,  <?. 

Solution. 

log(a-b)=    1.6928469 

a  =  234.7  Colog(a  +  b)  =    7.3766473 

b  =  185.4  p 

log  cot  ~  =  10.0409920 


a-b=    49.3 
a  +  b  =  420.1 

.-.  ^  =  42°  18'. 

2 


2 
A-B 


log  tan  t^—-=    9.1104862 


2 
A-B 


=  7°  20'  56". 


-"•      I      -P   A  f70    Af)\ 

2  log  b  =  2.2681097 

.-.  A  =  55°    2' 56",  log  sin  C  =  9.9980683 

B  =  40°  21'   4",  c°l°g  sin  B  =  0.1887804 

C  =  84°  36',  log  c  =  2.4549584 

c  =  285.0745. 


OBLIQUE   TRIANGLES.  177 

Ex.  2.  Given  a  =  .062387,  b  =  .023475,  C  =  110°  32' ;  find 
A,  B,  c. 

Ans.  A  =  52°  10'  33" ;  B  =  17°  17'  27" ;  c  =  .0739635. 

118.  Case  IV.   Given  the  three  sides,  as  a,  b,  c;  find  A,  B,  C. 

The  solution  in  this  case  may  be  performed  by  the  for- 
mulae of  Art.  99.  By  means  of  these  formulae  we  may 
compute  two  of  the  angles,  and  find  the  third  by  subtract- 
ing their  sum  from  180°.  But  in  practice  it  is  better  to 
compute  the  three  angles  independently,  and  check  the 
accuracy  of  the  work  by  taking  their  sum. 

If  only  one  angle  is  to  be  found,  the  formulae  for  the  sines 
or  cosines  may  be  used.  If  all  the  angles  are  to  be  found, 
the  tangent  formulas  are  the  most  convenient,  because  then 
we  require  only  the  logarithms  of  the  same  four  quantities, 
s}  s  —  a,  s  —  b}  s  —  c,  to  find  all  the  angles ;  whereas  the  sine 
and  cosine  formulae  require  in  addition  the  logs  of  a,  b,  c. 

The  tangent  formulae  (Art.  99)  may  be  reduced  as  follows : 


l(s  —  a)  (s  —  6)  (s  —  c) 

s  — 


2      s  —  a 


Similarly,          tan  —  =  —  ^— > 

2i      s  —  o 


(Art.  102) 


2      a-c 

NOTE,  —  The  quantity  r  is  the  radjqe  of  the  inscribed  circle  (Art.  102). 


1 

178  PLANE  TRIGONOMETRY. 


Ex.  1.    Given  a  =  13,  b  =  14,  c  =  15 ;  find  A,  B,  C. 

Solution. 

a  =  13  log(s-a)  =    .9030900 

6  =  14  log  («-&)  =    .8450980 

c  =  15  log(s-c)  =    .7781513 


2  s  =  42  colog  s  =  8.6777807 

s  =  21  log?-2  =  1.2041200 

logr=    .6020600. 


S  —  QJ  —  Oj 

8  _  b  =  7^  .-.log  tan  A  =  9.6989700. 

s  -  c  =  6'  .-.log  tan  -  =  9.7569620. 

.-.  log  tan  -  =  9.8239087. 

.-.  A  =  53°  7'  48".38  ; 
B  =  59°29f23'M8; 
C  =  67°  22'  48".44. 

Without  the  use  of  logarithms,  the  angles  may  be  found 
by  the  cosine  formulae  (Art.  96).  These  may  sometimes 
be  used  with  advantage,  when  the  given  lengths  of  a,  b,  c 
each  contain  less  than  three  digits. 

Ex.  2.  Find  the  greatest  angle  in  the  triangle  whose 
sides  are  13,  14,  15. 

Let  a  =  15,  b  =  14,  c  =  13.     Then  A  is  the  greatest  angle. 

™  A      62  +  c2-a2      142  +  132  -  152 

^T  2x14x13 

=  ^  =  .384615  =  cos  67°  23',  nearly 
(by  the  table  of  natural  sines). 

.-.  the  greatest  angle  is  67°  23'. 


EXAMPLES.  179 

EXAMPLES. 

1.  Given  a  =  254,  B  =  16°,  0  =  64°;  find  6  =  71.0919. 

2.  Given    c  =  338.65,    A  =  53°  24',   B  =  66°  27';    find 
a  =  313.46. 

.     3.   Given    c  =  38,    A  =  48°,    B  =  54°;    find    a  =  28.87, 
b  =  31.43. 

4.   Given    a  =  7012.5,    B  =  38°  12'  48",    C  =  60°;    find 
b  and  c.  ^ns.  5  =  4382.82  ;  c  =  6135.94. 


5.  Given  a  =  528,  b  =  252,  A  =  124°  34';  find  B  and  C. 

Ans.  B  =  23°8'33";  C  =  32°  17'  27". 

6.  Given  a  =  170.6,  6  =  140.5,  B  =  40°;  find  A  and  C. 

Ans.  A  =  51°  18'  21",  or  128°  41'  39"  ; 
C  =  88°  41'  39",  or    11°  18'  21". 

7.  Given  a  =  97,  6  =  119,  A  =  50°;  find  B  and  C. 

Ans.  B  =  70°    0'  56",  or  109°  59'    4"  ; 
C  =  59°  59'    4",  or    20°    0'  56". 

8.  Given  a  =  7,  6  =  8,  A  =  27°  47'  45"  ;  find  B,  C,  c. 

Ans.  B  =    32°  12'  15",  or  147°  47'  45"  ; 
C  =  120°,  or      4°  24'  30"  ; 

c  =  13,  or  1.15385. 

9.  Given  6  =  55,  c  =  45,  A  =  6°  ;  find  B  and  C. 

Ans.  B  =  149°  20'  31";  C  =  24°  39'  29". 

10.  Given  6  =  131,  c  =  72,  A  =  40°  ;  find  B  and  C. 

Ans.  B  =  108°  36'  30"  ;  C  =  31°  23'  30". 

11.  Given  a  =  35,  6  =  21,  C  =  50°  ;  find  A  and  B. 

Ans.  A  =  93°  11'  49";  B  =  36°  48'  11". 

12.  Given  a  =  601,  6  =  289,  C  =  100°  19'  6";  find  A  and  B. 

Ans.  A  =  56°  8'  42";  B  =  23°  32'  12", 


180  PLANE  TRIGONOMETRY. 

13.  Given  a  =  222,  b  =  318,  c  =  406 ;  find  A  =  32°  57'  8". 

14.  Given  a  =  275.35,  6=189.28,  c= 301.47  ;  find  A,  B,  C. 
Ans.  A  =  63°  30'  57" ;  B  =  37°  58'  20" ;  C  =  78°  30'  43". 

15.  Given  a  =  5238,  b  =  5662,  c  =  9384 ;  find  A  and  B. 

Ans.  A  =  29°  17'  16" ;  B  =  31°  55'  31". 

16.  Given  a  =  317,  b  =  533,  c  =  510 ;  find  A,  B,  C. 
Ans.  A  =  35°18'0";  B  =  76°  18' 52";  C  =  68°  23'  8". 

119.  Area  of  a  Triangle  (Art.  101). 

EXAMPLES. 
Find  the  area : 

1.  Given  a  =  116.082,  6  =  100,  C  =  118°  15' 41". 

Ans.  5112.25. 

2.  Gfiven  a  =  8,  6  =  5,  C  =  60°.  17.3205. 

3.  Given  6  =  21.5,  c  =  30.456,  A  =  41°  22'.         216.372. 

4.  Given  a  =  72.3,  A  =  52°  35',  B  =  63°  17'.       2644.94. 

5.  Given  6  =  100,  A  =  76°  38'  13",  C  =  40°  5'.   3506.815. 

6.  Given  a  =  31.325,  B  =  13°  57'  2",  A  =  53°  11'  18". 

Ans.  135.3545. 

7.  Given  a  =  .582,  6  =  .601,  c  =  .427.  .117655. 

8.  Given  a  =  408,  6  =  41,  c  =  401.  8160. 

9.  Given  a  =  .9,  6  =  1.2,  c  =  1.5.  .54. 

10.  Given  a  =  21,  6  =  20,  c  =  29.  210. 

11.  Given  a  =  24,  6  =  30,  c  =  18.  216. 

12.  Given  a  =  63.89,  6  =  138.24,  c  =  121.15.          3869.2. 


HEIGHT  OF  AN  OBJECT.  181 


MEASUREMENT    OF    HEIGHTS   AND    DISTANCES. 

120.  Definitions.  —  One  of  the  most  important  applica- 
tions of  Trigonometry  is  the  determination  of  the  heights 
and  distances  of  objects  which  cannot  be  actually  measured. 

The  actual  measurement,  with  scientific  accuracy,  of  a 
line  of  any  considerable  length,  is  a  very  long  and  difficult 
operation.  But  the  accurate  measurement  of  an  angle,  with 
proper  instruments,  can  be  made  with  comparative  ease  and 
rapidity. 

By  the  aid  of  the  Solution  of  Triangles  we  can  determine  : 

(1)  The  distance  between  points  which  are  inaccessible. 

(2)  The  magnitude  of  angles  which  cannot  be  practically 
observed. 

(3)  The   relative   heights   of    distant    and    inaccessible 
points. 

A  vertical  line  is  the  line  assumed  by  a  plummet  when 
freely  suspended  by  a  cord,  and  allowed  to  come  to  rest. 

A  vertical  plane  is  any  plane  containing  a  vertical  line. 

A  horizontal  plane  is  a  plane  perpendicular  to  a  vertical 
line. 

A  vertical  angle  is  one  lying  in  a  vertical  plane, 

A  horizontal  angle  is  one  lying  in  a  horizontal  plane. 

An  angle  of  elevation  is  a  vertical  angle  having  one  side 
horizontal  and  the  other  ascending. 

An  angle  of  depression  is  a  vertical  angle  having  one  side 
horizontal  and  the  other  descending. 

By  distance  is  meant  the  horizontal  distance,  unless  other- 
wise named. 

By  height  is  meant  the  vertical  height  above  or  below  the 
horizontal  plane  of  the  observer. 

For  a  description  of  the  requisite  instruments,  and  the 
method  of  using  them,  the  student  is  referred  to  books  on 
practical  surveying.* 

*  See  Johnson's  Surveying,  Gillespie's  Surveying,  Clarke's  Geodesy,  Gore's 
Geodesy,  etc. 


182  PLANE  TRIGONOMETRY. 

121.  To  find  the   Height  of   an   Object   standing  on  a 
Horizontal  Plane,  the  Base  of  the  Object 
being  Accessible. 

Let  BC  be  a  vertical  object,  such  as  /'' 

a  church  spire  or  a  tower. 

From  the  base  C  measure  a  horizon-   £ ~ 

tal  line  CA. 

At  the  point  A  measure  the  angle  of  elevation  CAB. 

We  can  then  determine  the  height  of  the  object  BC ;  for 

BC  =  AC  tan  CAB. 


EXAMPLES. 

1.  If  AC  =  100  feet  and  CAB  =  60°,  find  BC. 

Ans.  173.2  feet. 

2.  If  AC  =  125  feet  and  CAB  =  52°  34',  find  BC. 

Ans.  163.3  feet. 

3.  AC,  the  breadth  of  a  river,  is  100  feet.     At  the  point 
A,  on  one  bank,  the  angle  of  elevation  of  B,  the  top  of  a  tree 
on  the  other  bank  directly  opposite,  is  25°  37' ;  find  the 
height  of  the  tree.  Ans.  47.9  feet. 

122.  To  find  the  Height  and  Distance  of  an  Inaccessible 

Object  on  a  Horizontal  Plane. 

r. 

Let  CD  be  the  object,  whose  base  D  /; 

is  inaccessible ;  and  let  it  be  required  /'/ 

to  find  the  height  CD,  and  its  horizon-  ^'$    /  \a 

tal  distance  from  A,  the  nearest  acces-  B  ''  a  '  £ — ] — 
sible  point. 


(1)   At  A  in  the  horizontal  line  BAD  observe  the  Z  DAC 
=  a  ;  measure  AB  =  a,  and  at  B  observe  the  Z.  DBC  =  /?. 


Then  CA  =  (Art.  95) 

sin  (a  -  (3) 


HEIGHT  OF  AN  OBJECT.  183 

rD  —  pA  •      _  a  sin  a  sin  ft 

sin  (a  —  ft)  ' 

a  sin  8  cos  a 

and  AD  =  AC  cos  a  =  — — ?-    — •• 

sin  (a  —  /?) 

(2)    TF/ien  £/ie  Zwe  BA  cannot  be  measured  directly  toward 
the  object. 

At  A  observe  the  vertical  /.  CAD 
=  «,  and  the  horizontal  Z.  DAB  =  ft ; 
measure  AB  =  a,  and  at  B  observe 
the  Z.  DBA  =  y.  \^ 

Then  AD=sT^^-.  \-Z7 

.-.  CD  =  AD  tan  a 

_  a  sin  y  tan  a 


EXAMPLES. 

1.  A  river  300  feet  wide  runs  at  the  foot  of  a  tower, 
which  subtends  an  angle  of  22°  30'  at  the  edge  of  the  remote 
bank ;  find  the  height  of  the  tower.  Ans.  124.26  feet. 

2.  At  360  feet  from  the  foot  of  a  steeple  the  elevation  is 
half  what  it  is  at  135  feet ;  find  its  height.      Ans.  180  feet. 

3.  A  person  standing  on  the  bank  of  a  river  observes  the 
angle  subtended  by  a  tree  on  the  opposite  bank  to  be  60°, 
and  when  he  retires  40  feet  from  the  river's  bank  he  finds 
the  angle  to  be  30° ;   find  the  height  of  the  tree  and  the 
breadth  of  the  river.  Ans.  20 V3 ;  20. 

4.  What  is  the  height  of  a  hill  whose  angle  of  elevation, 
taken  at  the  bottom,  was  46°,  and  100  yards  farther  off,  on 
a  level  with  the  bottom,  the  angle  was  31°  ? 

Ans.  143.14  yards. 


184  PLANE  TRIGONOMETRY. 

123.  To  find  the  Height  of  an  Inaccessible  Object  situated 
above  a  Horizontal  Plane,  and  its 
Height  above  the  Plane. 

Let  CD  be  the  object,  and  let 
A  and  B  be  two  points  in  the 
horizontal  plane,  and  in  the  same 
vertical  plane  with  CD. 

At  A,  in  the  horizontal  line 
B  AE,  observe  the  A  C  AE  =  a, 

and   DAE  =  y;    measure    AB  =  a,    and   at   B   observe  the 
Z.  CBE  =  ft. 


Then  CE  =  (Art.  122) 

Bin  (a  -ft) 


Also,  AE  =  (Art.  122) 

sin  (a—  ft) 


DE  —  a  COS  a  s"1 


sin  (a  —  ft) 

...  CD=      asmfi     jsiiitt 
sin  (a-  ft)1 

—  a  sin/?  sin  (a  —  y) 

cos  y  sin  («  —  ft) 
) 

EXAMPLES. 

1.  A  man  6  feet  high  stands  at  a  distance  of  4  feet  9 
inches  from  a  lamp-post,  and  it  is  observed  that  his  shadow 
is  19  feet  long  :  find  the  height  of  the  lamp.      Ans.  1\  feet. 

2.  A  flagstaff,  25  feet  high,  stands  on  the  top  of  a  cliff, 
and  from  a  point  on  the  seashore  the  angles  of  elevation 
of  the  highest  and  lowest  points  of  the  flagstaff  are  observed 
to  be  47°  12'  and  45°  13'  respectively  :  find  the  height  of  the 
cliff.  Ans.  348  feet. 

3.  A   castle  standing  on  the  top  of  a  cliff  is  observed 
from  two  stations  at  sea,  which  are  in  line  with  it;  their 


HEIGHT  OF  AN  OBJECT.  185 

distance  is  a  quarter  of  a  mile:  the  elevation  of  the  top 
of  the  castle,  seen  from  the  remote  station,  is  16°  28';  the 
elevations  of  the  top  and  bottom,  seen  from  the  near 
station,  are  52°  24'  and  48°  38'  respectively:  (1)  what  is 
its  height,  and  (2)  what  its  elevation  above  the  sea  ? 

Ana.   (1)  60.82  feet ;   (2)  445.23  feet. 

124.  To  find  the  Distance  of  an  Object  on  a  Horizontal 
Plane,  from  Observations  made  at  Two  Points  in  the  Same 
Vertical  Line,  above  the  Plane. 


Let  the  points  of  observation  A 
and  B  be  in  the  same  vertical  line, 
and  at  a  given  distance  from  each 
other  ;  let  C  be  the  point  observed, 

whose  horizontal  distance  CD  and   ^ 

P  D 

vertical  distance  AD  are  required. 

Measure  the  angles  of  depression,  6BC,  aAC,  equal  to  a 
and  (3  respectively,  and  denote  AB  by  a. 

Then  BD  =  CD  tan  «,  AD  =  CD  tan/?. 

.-.  a  =  CD  (tan  a  —  tan  0). 
p-pj  _  a  cos  a  cos  ft 
~  sin  (a  -  (3)  ' 

and 


sin  (a  —  (3) 
EXAMPLES. 

1.  From  the  top  of  a  house,  and  from  a  window  30  feet 
below  the  top,  the  angles  of  depression  of  an  object  on  the 
ground  are  15°  40'  and  10°  :  find  (1)  the  horizontal  distance 
of  the  object,  and  (2)  the  height  of  the  house. 

AIM.  l  (1)  288.1  feet  ;   (2)  80.8  feet. 

2.  From  the  top  and  bottom  of  a  castle,  which  is  68  feet 
high,  the  depressions  of  a  ship  at  sea  are  observed  to  be 
16°  28'  and  14°  :  find  its  distance.  Ans.  570.2  yards. 


186  PLANE  TRIGONOMETRY. 

125.  To  find  the  Distant  between  Two  Inaccessible  Objects 
on  a  Horizontal  Plane. 

Let  C  and  D  be  the  two  inac-  __- x7 

cessible  objects. 

Measure  a  base  line  AB,  from 
whose  extremities  C  and  D  are 
visible.  At  A  observe  the  angles 
CAD,  DAB;  and  at  B  observe 
the  angles  CBA  and  CBD. 

Then,  in  the  triangle  ABC,  we  know  two  angles  and  the 
side  AB.  .-.  AC  may  be  found.  In  the  triangle  ABD  we 
know  two  angles  and  the  side  AB.  .•.  AD  may  be  found. 

Lastly,  in  the  triangle  ACD,  AC  and  AD  have  been  deter- 
mined, and  the  included  angle  CAD  has  been  measured ; 
and  thus  CD  can  be  found. 

EXAMPLES. 

1.  Let  AB  =  1000  yards,  the  angles  BAG,  BAD  =  76°  30' 
and  44°  10',   respectively ;    and  the  angles  ABD,    ABC  = 
81°  12'  and  46°  5',  respectively :  find  the  distance  between 
C  and  D.  Ans.  669.8  yards. 

2.  A  and  B  are  two  trees  on  one  side  of  a  river ;  at  two 
stations  P  and  Q  on  the  other  side  observations  are  taken, 
and  it  is  found  that  the  angles  APB,  BPQ,  AQP  are  each 
equal  to  30°,  and  that  the  angle  AQB  is  equal  to  60°.     If 
PQ  =  a,  show  that 

AB  =  -V21. 
6 

126.  The  Dip  of  the  Horizon.  —  Since  the  surface  of  the 
earth  is  spherical,  it  is  obvious  that  an  object  on  it  will  be 
visible  only  for  a  certain  distance  depending  on  its  height ; 
and,  conversely,  that  at  a  certain  height  above  the  ground 
the  visible  horizon  will  be  limited. 


THE  DIP   OF  THE  HORIZON. 


187 


Let  0  be  the  centre  of  the  earth,  P  a  point  above  the  sur- 
face, PD  a  tangent  to  the  surface  at  c 
D.     Then  D  is  a  point  on  the  terres- 
trial horizon ;  and  CPD,  which  is  the 
angle  of  depression  of  the  most  dis- 
tant point  on  the  horizon  seen  from 
P,  is  called  the  dip  of  the  horizon  at  P. 
The  angle  DOP  is  equal  to  it. 

Denote  the  angle  CPD  by  0,  the 
height  AP  by  h,  and  the  radius  OD 
by  r.  Then 

h  =  OP  -  OA  =  r  sec  0  -  r 

=  r(l  —  cos  0) 
cos  0 


r  = 


hcos  6 


1-COS0 


rtan0  =  - 


h  sin  0 


Also 


1-COS0 

0 

2     '; 

PD2  =  PA  x  PB  =  h(h  +  2r) 


(Art.  48,  Ex.  8) 
.     .     (Geom.) 


Since,  in  all  cases  which  can  occur  in  practice,  h  is  very 
small  compared  with  2r,  we  have  approximately 


Let  n  =  the  number  *  of  miles  in  PD,  h  =  the  feet  in  PA, 
and  r  =  4000  miles  nearly.     Then 

A==PD^=     (5280  n)2 
=  2r       8000  x  5280 


5280  n2=  5.28 
8000   =      8 


*  It  will  be  noticed  that  n  is  a  number  merely,  and  that  the  result  will  be  in  feet, 
since  the  miles  have  been  reduced  to  feet. 


188 


PLANE  TRIGONOMETRY. 


That  is,  the  height  at  which  objects  can  be  seen  varies  as 
the  square  of  the  distance. 

Thus,  if  ?i  =  1  mile,  we  have 

h  =  |  feet  =  8  inches  ; 
if  71  =  2  miles,  h  =  f  -  22  =  §  feet,  etc.,  etc. 

Thus  it  appears  that  an  object  less  than  8  inches  above 
the  surface  of  still  water  will  be  invisible  to  an  eye  on  the 
surface  at  the  distance  of  a  mile. 

Example.  From  a  balloon,  at  an  elevation  of  4  miles,  the 
dip  of  the  sea-horizon  is  observed  to  be  2°  33'  40"  :  find  (1) 
the  diameter  of  the  earth,  and  (2)  the  distance  of  the 
horizon  from  the  balloon. 

Ans.   (1)  8001.24  miles ;   (2)  178.944  miles. 

127.  Problem  of  Pothenot  or  of  Snellius.  —  To  determine 
a  point  in  the  plane  of  a  given  triangle,  at  which  the  sides  of 
the  triangle  subtend  given  angles. 

Let  ABC  be  the  given  triangle,  and 
P  the  required  point.  Join  P  with 
A,B,  C. 

Let  the  given  angles  APC,  BPC  be 
denoted  by  a,  ft,  and  the  unknown 
angles  PAC,  PBC  by  x,  y  respectively ; 
then  a  and  ft  are  known ;  and  when  x 
and  y  are  found,  the  position  of  P  can 
be  determined,  for  the  distances  PA 
and  PB  can  be  found  by  solving  the 
triangles  PAC,  PBC. 

We  have    x  +  y  =  2ir  —  a  —  ft  —  C      . 
Also          &sinx=osiM==pa 

sin  a         sin  ft 

Assume  an  auxiliary  angle  <£  such  that 
a  sin  a 


(1) 


tan  $  = 
then  the  value  of 


. 

o  sin  ft 

can  be  found  from  the  tables. 


EXAMPLES.  189 


mr  Sin  X 

Thus,  —  =  tan  <f>. 

smy 


=  tan  (0  -  45°) 


sm  ic  +  sin  i/      tan  <£  -f  1 

[(14)  of  Art.  61]. 

.-.  tan  %(x  —  y)  =  tan  \  (x  +  y)  tan  (<£  —  45°) 

[(13)  of  Art.  61]. 

=  tan  (45°  -  <£)  tan  \  («  +  ft  +  C)     .     (2) 
thus  from  (1)  and  (2)  a;  and  #  are  found. 

EXAMPLES. 

Solve  the  following  right  triangles  : 

1.  Given     a=51.303,  c=150; 

find      A  =  20°,  B  =  70°,  6  =  140.95. 

2.  Given     a=157.33,  c=250; 

find     A  =  39°,  B  =  51°,  6  =  194.28. 

3.  Given     a  =104,  c=185; 

find     A  =  34°12'19".6,  B  =  55°47'40".4,    6  =  153. 

4.  Given     a=304,  c=425; 

find      A=45°40'  2".3,  B=  44°  19'  57  ".7,    6  =  297. 

5.  Given     6=3,  c=5; 

find      A=53°   7'48".4,  B=36°52'll".6,    a=4. 

6.  Given     6=15,  c=17; 

find      A=28°  4'20".9,  B  =  61°55'39".l,    a=8. 

7.  Given     6=21,  c=29; 

find     A=43°36'10".l,  B  =  46°23'49'f.9,    a=20. 

8.  Given     6=7,  c=25; 

find     A=73°44'23".3,  B  =  16°  15'  36".  7,    a=24. 


190  PLANE  TRIGONOMETRY. 

9.    Given     6=33,  c=65; 

find  A=59°29'23".2,  B  =  30°30'36".8,    a=56. 

10.  Given      c=625,  A =44°; 

find  o=434.161,  6=449.587. 

11.  Given      c=300,  A=52°; 

find  a=236.403,  6=184.698. 

12.  Given     c=13,  A=   67°22'48".5; 

find  B  =  22°37'll".5,  a  =  12  6=5. 

13.  Given  A  =  77°19'10".6,  c=41; 

find  B  =  12°40'49".4,  a=40,  6  =  9. 

14.  Given  B  =  48°53'16".5,  c=73; 

find  A=41°   6'43".5,  a=48,  6  =  55. 

15.  Given  B  =  64°  0'38".8,  c=89; 

find  A=25°59'21".2,  a=39,  6=80. 

16.  Given  A  =  77°19'10".6,  a=40; 

find  B  =  12°40'49".4,  6=9,  c=41. 

17.  Given  A=87°12'20".3,  a=840; 

find  B=  2°47'39".7,  6=41,  c=841. 

18.  Given  A=32°31'13".5,  a=336; 

find  B  =  57°28'46".5,  6=527,  c=625. 

19.  Given  A=82°41'44",  a=1100; 

find  B=  7°18'16",  6=141,  c=1109. 

20.  Given  A=75°23'18".5,  6=195; 

find  B  =  14°36'41f'.5,  a=748,  c=773. 

21.  Given  B  =  87°  49' 10",  6  =  42536.37; 

find  A=  2°10f50",  a=1619.626,          c=42567.2. 


EXAMPLES. 


191 


22.  Given  A=  88°59'  6=2.234875; 

find    B=     1°   1',  a=  125.9365,          c=  125.9563. 

23.  Given  A=  35°  16' 25",      a  =  388.2647; 

find    B=  54°43'35,"      6  =  548.9018,          c=672.3412. 

24.  Given  a=   7694.5,  6  =  8471; 

find    A=  42°  15',  B  =  47°45',  c= 11444. 

25.  Given  a  =  736,  6  =  273; 

find    A=  69°38'56".3,  B  =  20°21'3".7,       c=785. 

26.  Given  a=200,  6  =  609; 

find   A=   18°  10' 50",     B  =  71°49'10",        c=641. 

27.  Given  a=276,  6  =  493; 

find   A=  29°14'30".3,  B  =  60° 45 '29. "7,     c=565. 

28.  Given  a=  396,  6=403; 

find   A=  44° 29' 53",      B=  45°30'   7",      c=565. 

29.  Given  a=278.3,  6=314.6; 

find   A=  41° 30',  B=  48°30',  c=420. 

30.  Given  a=372,  6=423.924; 

find   A=  41°  16'  2".7,  B=  48°43'57".3,  c=564. 

31.  Given  a=526.2,  6=414.745; 

find   A=  51°45'18".7,  B=  38°14'41".3,  c=670. 

Solve  the  following  oblique  triangles  : 

32.  Given  B=  50°  30',  C  =  122°   9',  a  =  90; 

find   A=     7° 21',  6=542.850,  c=595.638. 

33.  Given  A=  82°20',  B=  43°20',  a=479; 

find    C  =  54°20',  6=331.657,  c=392.473. 

34.  Given  A=   79° 59',  B=  44° 41',  a=795; 

find    C=  51° 20',  6=567.888,  c=663.986. 


192 


PLANE   TRIG  ON  OMETR  Y. 


35. 

Given 

B  = 

37° 

58', 

C=  65°  2', 

a=999; 

find 

A  = 

77° 

0', 

6  =  630.< 

r71, 

c=829.480. 

36. 

Given 

A  = 

70° 

55', 

C=  52° 

9', 

a  =  6412; 

find 

B  = 

56° 

56', 

6  =  5686.00, 

c=5357.50. 

37. 

Given 

A  = 

48° 

20', 

B=  81° 

or  -\  (\n         7,       x  7^  . 
—    -LD   ,        0  =  O.  »O  ; 

find 

C  = 

50° 

37' 

44", 

a=4.3485, 

o=4.5. 

38. 

Given 

A= 

72° 

4', 

B=  41° 

56' 

18", 

c=24; 

find 

C  = 

65° 

59' 

42", 

a  =  24.995, 

6=17.559. 

39. 

Given 

A  = 

43° 

36' 

10".l, 

0=124° 

58' 

33".6, 

6=29; 

find 

B  = 

11° 

25' 

16".3, 

a  =101, 

c=120. 

40. 

Given 

A  = 

69° 

59' 

2".5, 

C=   70° 

42' 

30", 

6  =  149; 

find 

B  = 

39°  18' 

27".5, 

a=221, 

o=222. 

41. 

Given 

A= 

21° 

14' 

25", 

a  =345, 

6  =  695; 

find 

B  = 

46° 

52' 

10", 

r\  111° 

53' 

25", 

c=  883.65. 

or 

B'=133° 

7' 

50", 

C'=  25° 

37' 

45", 

o'=411.92. 

42. 

Given 

A  = 

41° 

13' 

0", 

a  =77.04* 

6  =  91.06; 

find 

B  = 

51° 

9' 

6", 

C=  87° 

37' 

54", 

c  =  116.82, 

or 

B'  = 

128° 

50'  54", 

C'=     9° 

56' 

6", 

o'  =  20.172. 

43. 

Given 

A= 

21° 

14' 

25", 

a=309, 

6=360; 

find 

B  = 

24° 

51' 

54", 

C  =  133°47' 

41", 

c=615.67, 

or 

B'= 

155° 

2' 

6", 

C'=     3° 

43' 

29", 

c'=55.41. 

44. 

Given 

B  = 

68° 

10'  24", 

a=83.856, 

6  =  83.153; 

find 

A= 

65° 

5' 

10", 

C=  45° 

44' 

26", 

c=65.696. 

45. 

Given 

B  = 

60° 

0' 

32", 

a=27.548,    " 

6  =  35.055; 

find 

A= 

42°  53' 

34", 

C=   77° 

5' 

54", 

c=39.453. 

46. 

Given 

A= 

60°, 

a  =120, 

6  =  80; 

find 

B  = 

35°  15' 

52", 

C=  84° 

44' 

8", 

c=137.9796. 

EXAMPLES.  193 

47.  Given  A=   50°,  «  =  119,  6=97; 

find    B  =  38°  38' 24",  C=  91°  21' 36",  c= 166.3. 

48.  Given  C=   65° 59',  a  =  25,  c=24; 

find    A=   72°   4' 48",  B=  41° 56' 12",  6=17.56, 

or      A'=107°55'12",  B'=     6°   6' 48". 

49.  Given  A  =  18°55'28".7,  a=13,  6  =  37; 

find    B=  67°22'48".l,  or  Bf  =  112°37'11".9. 

50.  Given  C=   15°11'21",  a=232,  6  =  229; 

find    A=  85°  11' 68",  B=   79°36'40",  c=61. 

51.  Given  C=126°12'14",  a=5132,  6=3476; 

find    A=  32°28'19",  B=  21°19'27",  c=7713.3. 

52.  Given  C=  55°12'  3",  a  =  20.71,  6  =  18.87; 

find    A=  67°28'51".5,  B=  57°19'5".5,  c=18.41. 

53.  Given  C=   12°35'  8",  a  =8.54,  6  =  6.39; 

find    A=136°15'48",  B=  31°   9'   4",  c=2.69. 

54.  Given  C=  34°   9' 16",  a=3184,  6=917; 

find    A=133°51'34",  B=   11° 59' 10",  c=2479.2. 

55.  Given  C=  32°10'63".8,  a=101,  6  =  29; 

find    A=136°23'49".9,  B=   11°25'16".3,  o=78. 

56.  Given.  C=  96°57'20".l,  a=401,  6=41; 

find    A=   77°19'10".6,  B=  5°43'29".2,  c=408. 

57.  Given  C=   30°40'35",  a=221,-  6=149; 

find    A  =  110°  0'57".5,  B=  39°18'27".5,  c=120. 

58.  Given  C=  66°59'25".4,  a =109,  6=61; 

find    A=   79°36'40",  B=   33°23'54".6,  c=102. 

59.  Given  C  =  131°24'44",  a  =  229,  6  =  109; 

find    A=  33°23'54".6,  B=   15°llf21".4,  c=312. 


194  PLANE   TRIGONOMETRY. 

60.  Given  0  =  104°  3' 51",     a =241,  6=169; 

find  A=45°46'16".5,B  =  30°  9'52".5,   c=332.97. 

61.  Given  a=289,  6=601,  c=712 ; 

find  A=23°32'12",    B  =  56°   8'42",     C  =  100°19f  6". 

62.  Given  a=17,  6=113,  c=120; 

findA=  7°37'42",     B  =  61°55'38",     C  =  110°26'40". 

63.  Given  a=15.47,  6=17.39,  c  =  22.88; 

find  A=42°30'44",     B  =  49°25'49",     C  =  88°3'27". 

64.  Given  a= 5134,  6=7268,  c=9313; 

find  A= 33°  15' 39",    B  =  50°56'  0",     C  =  95°48'21". 

65.  Given  a=99,  6  =  101,  c=158; 

find  A=37°22'19",     B  =  38°15'41",     C  =  104°22'0". 

66.  Given  a=ll,  6=13,  c=16; 

findA  =  43°  2' 56",     B  =  53°46r44",     C  =  83°10'20". 

67.  Given  a=25,  6=26,  c=27; 

findA=56°15'  4",    B =59° 51' 10",     C  =  63°53'46". 

68.  Given  a=197,  6  =  53,  c=240 ; 

find  A  =  31°53'26".8,B=  8°10'16".4,  C  =  139°56'16".8. 

69.  Given  a  =  509,  6=221,  c=480 ; 

find  A=84°32'50".5,B=25°36'30".7,  C=69°50'38".8. 

70.  Given  a=533,  6=317,  c  =  510 ; 

find  A  =  76°  18' 52",     B=35°18'  0".9,  C  =  68°23'7".l. 

71.  Given  a=565,  6=445,  c=606 ; 

find  A=62°51'32".9,  B  =  44°29'53",     C  =  72°38'34".l. 

72.  Given  a=10,  6=12,  c=14; 

find  A=44°24'55".2,B=57°  7'18",    C  =  78°27'47". 

73.  Given  a  =  .8706,  6  =  .0916,  c=.7902; 

find  A=149°49'0".4,B=  3°   1'56".2,  C  =  27°9'3".4. 


EXAMPLES.  195 

Find  the  area : 

74.  Given  o=10,      6  =  12,        C  =  60°.  Ans.  30V3. 

75.  "  a =40,        6  =  60,        C  =  30°.  600. 

76.  "  6  =  7,  c=5V2,   A  =  135°.                           17£. 

77.  «  a=32.5,     6  =  56.3,     C=47°5'30".  670. 

78.  «  6  =  149,     A=  70°42'30",  B  =  39°18'28".   15540. 

79.  «  c=8.025,  B  =  100°   5' 23",  C  =  31°  6' 12".  46.177. 

80.  «  a=5,  6=6,         c=7.                                   12. 

81.  «  a=625,      6=505,      c=904.  151872. 

82.  «  a=409,      6  =  169,      c=510.  30600. 

83.  «  a=577,       6=73,        c=520.  12480. 

84.  «  a  =  52.53,    6  =  48.76,  c=44.98.  1016.9487. 

85.  «  a=13,        6  =  14,        c=15.  84. 

86.  "     a=242  yards,  6  =  1212  yards,  c=1450  yards. 

Ans.  6  acres. 

87.  «    a=7.152,    6=8.263,  c=9.375.  28.47717. 

88.  The  sides  of  a  triangle  are  as  2  :  3  :  4  :  show  that  the 
radii  of  the  escribed  circles  are  as  1 :  i  :  1. 

O         o 

89.  The  area  of  a  triangle  is  an  acre ;  two  of  its  sides 
are  127  yards  and  150  yards :  find  the  angle  between  them. 

Ans.  30°  32' 23". 

90.  The  adjacent  sides  of  a  parallelogram  are  5  and  8, 
and  they  include  an  angle  of  60° :  find  (1)  the  two  diag- 
onals, and  (2)  the  area.        Ans.   (1)  7,  Vl29 ;    (2)  20  V3. 

91.  Two  angles  of  a  triangular  field  are  22£°  and  45°,  and 
the  length  of  the  side  opposite  the  latter  is  a  furlong.     Show 
that  the  field  contains  2    acres. 


196  PLANE  TRIGONOMETRY. 


HEIGHTS  AND  DISTANCES. 

92.  At  a  point  200  feet  in  a  horizontal  line  from  the  foot 
of  a  tower,  the  angle  of  elevation  of  the  top  of  the  tower  is 
observed  to  be  60° :  find  the  height  of  the  tower. 

Ans.  346  feet. 

93.  From  the  top  of  a  vertical  cliff,  the  angle  of  depres- 
sion of  a  point  on  the  shore  150  feet  from  the  base  of  the 
cliff,  is  observed  to  be  30° :  find  the  height  of  the  cliff. 

Ans.  86.6  feet. 

94.  From  the  top  of  a  tower  117  feet  high,  the  angle  of 
depression  of  the  top  of  a  house  37  feet  high  is  observed  to 
be  30° :  how  far  is  the  top  of  the  house  from  the  tower  ? 

Ans.  138.5  feet. ' 

95.  The  shadow  of  a  tower  in  the  sunlight  is  observed 
to  be  100  feet  long,  and  at  the  same  time  the  shadow  of  a 
lamp-post  9  feet  high  is  observed  to  be  3  V3  feet  long :  find 
the  angle  of  elevation  of  the  sun,  and  height  of  the  tower. 

Ans.  60°;   173.2  feet. 

96.  A  flagstaff  25  feet   high  stands  on  the   top   of    a 
house;    from  a  point  on  the  plain   on  which   the   house 
stands,  the  angles  of  elevation  of  the  top  and  bottom  of 
the  flagstaff  are  observed  to  be  60°  and  45°  respectively : 
find  the  height  of  the  house  above  the  point  of  observation. 

Ans.  34.15  feet. 

97.  From  the  top  of  a  cliff  100  feet  high,  the  angles  of 
depression  of  two  ships  at  sea  are  observed  to  be  45°  and 
30°  respectively  ;  if  the  line  joining  the  ships  points  directly 
to  the  foot  of  the  cliff,  find  the  distance  between  the  ships. 

Ans.  73.2. 

98.  A  tower  100  feet  high  stands  on  the  top  of  a  cliff ; 
from  a  point  on  the  sand  at  the  foot  of  the  cliff  the  angles 


EXAMPLES.  197 

of  elevation  of  the  top  and  bottom  of  the  tower  are  observed 
to  be  75°  and  60°  respectively :  find  the  height  of  the  cliff. 

Ans.  86.6  feet. 

99.  A  man  walking  a  straight  road  observes  at  one  mile- 
stone a  house  in  a  direction  making  an  angle  of  30°  with 
the  road,  and  at  the  next  milestone  the  angle  is  60° :  how 
far  is  the  house- from  the  road  ?  Ans.  1524  yds. 

100.  A  man  stands  at  a  point  A  on  the  bank  AB  of  a 
straight  river  and  observes  that  the  line  joining  A  to  a  post 
C  on  the  opposite  bank  makes  with  AB  an  angle  of  30°. 
He  then  goes  400  yards  along  the  bank  to  B  and  finds  that 
BC  makes  with  BA  an  angle  of  60° :  find  the  breadth  of 
the  river.  Ans.  173.2  yards. 

101.  From  the  top  of  a  hill  the  angles  of  depression  of 
the  top  and  bottom  of  a  flagstaff  25  feet  high  at  the  foot 
of  the  hill  are  observed  to  be  45°  13'  and  47°  12'  respectively  : 
find  the  height  of  the  hill.  Ans.  373  feet. 

102.  From  each  of  two  stations,  east  and  west  of  each 
other,  the  altitude  of  a  balloon  is  observed  to  be  45°,  and 
its  bearings  to  be  respectively  N.W.  and  KE. ;  if  the  sta- 
tions be  1  mile  apart,  find  the  height  of  the  balloon. 

Ans.  3733  feet. 

103.  The  angle  of  elevation  of  a  balloon  from  a  station 
due  south  of  it  is  60°,  and  from  another  station  due  west 
of  the  former  and  distant  a  mile  from  it  is  45° :  find  the 
height  of  the  balloon.  Ans.  6468  feet. 

104.  Find  the  height  of  a  hill,  the  angle  of  elevation  at 
its  foot  being  60°,  and  at  a  point  500  yards  from  the  foot 
along  a  horizontal  plane  30°.  Ans.  250  V3  yards. 

105.  A  tower  51  feet  high  has  a  mark  at  a  height  of  25 
feet  from  the  ground :  find  at  what  distance  from  the  foot 
the  two  parts  subtend  equal  angles. 


198  PLANE  TRIGONOMETRY. 

106.  The  angles  of  a  triangle  are  as  1  :  2  :  3,  and  the  per- 
pendicular from  the  greatest  angle  on  the  opposite  side  is 
30  yards  :  find  the  sides.  Ans.  20V3,  60,  40  V3. 

107.  At  two  points  A,  B,  an  object  DE,  situated  in  the 
same  vertical  line  CE,  subtends  the  same  angle  a ;  if  AC, 
BC  be  in  the  same  right  line,  and  equal  to  a  and  6,  respec- 
tively, prove 

DE  =  (a  +  1))  tan  «. 

108.  From  a  station  B  at  the  foot  of  an  inclined  plane 
BC  the  angle  of  elevation  of  the  summit  A  of  a  mountain  is 
60°,  the  inclination  of  BC  is  30°,  the  angle  BCA  135°,  and 
the  length  of  BC  is  1000  yards  :  find  the  height  of  A  over  B. 

Ans.  500(3  +  V3)  yards. 

109.  A  right  triangle  rests  on  its  hypotenuse,  the  length 
of  which  is   100  feet;  one  of  the   angles  is   36°,  and  the 
inclination  of  the  plane  of  the  triangle  to  the  horizon  is 
60°:    find  the  height  of  the  vertex  above  the  ground. 

Ans.  25  V3  cos  18°. 

110.  A  station  at  A  is  due  west  of  a  railway  train  at  B ; 
after  traveling  N.W.  6  miles,  the   bearing  of  A  from  the 
train  is  S.  221°  W. :  find  the  distance  AB.         Ans.  6  miles. 

111.  The  angles  of  depression  of  the  top  and  bottom  of  a 
column  observed  from  a  tower  108  feet  high  are  30°  and  60° 
respectively:  find  the  height  of  the  column.      Ans.  72  feet. 

112.  At  the  foot  of  a  mountain  the  elevation  of  its  sum- 
mit is  found  to  be  45°.     After  ascending  for  one  mile,  at  a 
slope  of  15°,  towards  the  summit,  its  elevation  is  found  to 
be  60° :  find  the  height  of  the  mountain. 

Ans.  — — -        miles. 

V2 

113.  A  and  B  are  two  stations  on  a  hillside.    The  inclina- 
tion of  the  hill  to  the  horizon  is  30°.     The  distance  between 
A  and  B  is  500  yards.     C  is  the  summit  of  another  hill  in 


EXAMPLES.  199 

the  same  vertical  plane  as  A  and  B,  on  a  level  with  A,  but 
at  B  its  elevation  above  the  horizon  is  15° :  find  the  distance 
between  A  and  C.  Ans.  500  (  V3  +  1) . 

114.  From  the  top  of  a  cliff  the  angles  of  depression  of 
the  top  and  bottom  of  a  lighthouse  97.25  feet  high  are 
observed  to  be  23°  17'  and  24°  19'  respectively  :  how  much 
higher  is  the  cliff  than  the  lighthouse  ?  Ans.  1942  feet. 

115.  The  angle  of  elevation  of  a  balloon  from  a  station 
due  south  of  it  is  47°  18'  30",  and  from  another  station  due 
west  of  the  former,   and  distant  671.38    feet  from  it,  the 
elevation  is  41°  14' :  find  the  height  of  the  balloon. 

Ans.  1000  feet. 

116.  A  person  standing  on  the  bank  of  a  river  observes 
the  elevation  of  the  top  of  a  tree  on  the  opposite  bank  to  be 
51°;  and  when  he  retires  30  feet  from  the  river's  bank  he 
observes  the  elevation  to  be  46° :  find  the  breadth  of  the 
river.  Ans.  155.823  feet. 

117.  From  the  top  of  a  hill  I  observe  two  milestones  on 
the  level  ground  in  a  straight  line  before  me,  and  I  find 
their  angles  of  depression  to  be  respectively  5°  and  15° : 
find  the  height  of  the  hill.  Ans.  228.6307  yards. 

118.  A  tower  is  situated  on  the  top  of  a  hill  whose  angle 
of  inclination  to  the  horizon  is  30°.     The  angle  subtended 
by  the  tower  at  the  foot  of  the  hill  is  found  by  an  observer 
to  be  15° ;  and  on  ascending  485  feet  up  the  hill  the  tower 
is  found  to  subtend  an  angle  of  30° :  find  (1)  the  height  of 
the  tower,  and  (2)  the  distance  of  its  base  from  the  foot  of 
the  hill.  Ans.   (1)  280.015 ;   (2)  765.015  feet. 

119.  The  angle  of  elevation  of  a  tower  at  a  place  A  due 
south  of  it  is  30° ;  and  at  a  place  B,  due  west  of  A,  and  at 
a  distance  a  from  it,  the  elevation  is  18° :   show  that  the 

height  of  the  tower  is  - 


200  PLANE  TRIGONOMETRY. 

120.  On  the  bank  of  a  river  there  is  a  column  200  feet 
high  supporting  a  statue  30  feet  high.     The  statue  to  an 
observer  on  the  opposite  bank  subtends  an  equal  angle  with 
a  man  6  feet  high  standing  at  the  base  of  the  column :  find 
the  breadth  of  the  river.  Ans.  10  V115  feet. 

121.  A  man  walking  along  a  straight  road  at  the  rate  of 
3  miles  an  hour,  sees  in  front  of  him,  at  an  elevation  of  60°, 
a  balloon  which  is  travelling  horizontally  in  the  same  direc- 
tion at  the  rate  of  6  miles  an  hour;  ten  minutes  after  he 
observes  that  the  elevation  is  30° :  prove  that  the  height  of 
the  balloon  above  the  road  is  440  V3  yards. 

122.  An  observer  in   a   balloon   observes   the   angle   of 
depression  of  an  object  on  the  ground,   due  south,  to   be 
35°  30'.     The  balloon  drifts  due  east,  at  the  same  elevation, 
for  2£   miles,  when  the  angle  of  depression  of  the  same 
object  is  observed  to  be   23°  14' :    find  the  height  of  the 
balloon.  Ans.  1.34394  miles. 

123.  A  column,  on  a  pedestal  20  feet  high,  subtends  an 
angle  45°  to  a  person  on  the  ground ;  on  approaching  20 
feet,  it  again  subtends  an  angle  45° :  show  that  the  height 
of  the  column  is  100  feet. 

124.  A  tower  51  feet  high  has  a  mark  25  feet  from  the 
ground :  find  at  what  distance  the  two  parts  subtend  equal 
angles  to  an  eye  5  feet  from  the  ground.  Ans.  160  feet. 

/  125.  From  the  extremities  of  a  sea-wall,  300  feet  long, 
the  bearings  of  a  boat  at  sea  were  observed  to  be  N.  23°  30' 
E.,  and  N.  35°  15'  W. :  find  the  distance  of  the  boat  from 
the  sea-wall.  Ans.  262.82  feet. 

126.  ABC  is  a  triangle  on  a  horizontal  plane,  on  which 
stands  a  'tower  CD,  whose  elevation  at  A  is  50°  3'  2" ;  AB  is 
100.62  feet,  and  BC  and  AC  make  with  AB  angles  40°  35'  17" 
and  9°  59'  50"  respectively  :  find  CD.  Ans.  101.166  feet. 


/ 


EXAMPLES.  201 

127.  The  angle  of  elevation  of  a  tower  at  a  distance  of 
20  yards  from  its  foot  is  three  times  as  great  as  the  angle 
of  elevation  100  yards  from  the  same  point  :  show  that  the 

300 
height  of  the  tower  is  -  -  feet. 

V7  . 

128.  A  man  standing  at  a  point  A,  due  south  of  a  tower 
built  on  a  horizontal  plain  ,  observes  the  altitude  of  the  tower 
to  be  60°.     He  then  walks  to  a  point  B  due  west  from  A 
and  observes  the  altitude  to  be  45°,  and  then  at  the  point  C 
in  AB  produced  he  observes  the  altitude  to  be  30°  :  prove 
that  AB  =  BC. 

v-i 

129.  The  angle  of  elevation  of  a  balloon,  which  is  ascend- 

ing  uniformly  and  vertically,  when  it  is  one  mile  high  is 
observed  to  be  35°  20'  ;  20  minutes  later  the  elevation  is 
observed  to  be  55°  40'  :  how  fast  is  the  balloon  moving  ? 

Ans.  3  (sin  20°  20')  (sec  55°  40')  (cosec  35°  20')  miles  per  hour. 

130.  The  angle  of  elevation  of  the  top  of  a  steeple  at  a 
place  due  south  of  it  is  45°,  and  at  another  place  due  west 
of  the   former   station  and  distant   100  feet  from  it  the 
elevation  is  15°  :   show  that  the  height  of  the  steeple  is 
60(3*  -3"*)  feet. 

131.  A  tower  stands  at  the  foot  of  an  inclined  plane 
whose  inclination  to  the  horizon  is  9°;  aline  is  measured 
up  the  incline  from  the  foot  of  the  tower,  of  100  feet  in 
length.     At  the  upper  extremity  of  this  line  the  tower  sub- 
tends an  angle  of  54°  :  find  the  height  of  the  tower. 

Ans.  114.4  feet. 

132.  The  altitude  of  a  certain  rock  is  observed  to  be  47°, 
and  after  walking  1000  feet  towards  the  rock,  up  a  slope 
inclined  at  an  angle  of  32°  to  the  horizon,  the  observer  finds 
that  the  altitude  is  77°  :  prove  that  the  vertical  height  of 
the  rock  above  the  first  point  of  observation  is  1034  feet. 


202  PLANE  TRIGONOMETRY. 

133.  From  a  window  it  is  observed  that  the  angle  of 
elevation  of  the  top  of  a  house  on  the  opposite  side  of  the 
street  is  29°,  and  the  angle  of  depression  of  the  bottom  of 
the  house  is  56°:  find  the  height  of  the  house,  supposing 
the  breadth  of  the  street  to  be  80  feet.         Ans.  162.95  feet. 

134.  A  and  B  are  two  positions  on  opposite  sides  of  a 
mountain ;  C  is  a  point  visible  from  A  and  B ;  AC  and  BC 
are  10  miles  and  8  miles  respectively,  and  the  angle  BCA  is 
60° :    prove  that  the  distance  between  A  and  B  is  9.165 
miles. 

135.  P  and  Q  are  two  inaccessible  objects  ;  a  straight  line 
AB,  in  the  same  plane  as  P  and  Q,  is  measured,  and  found 
to  be  280  yards ;  the  angle  PAB  is  95°,  the  angle  QAB  is 
47 J°,  the  angle  QBA  is  110°,  and  the  angle  PBA  is  52°  2Qf : 
find  the  length  of  PQ.  Ans.  509.77  yards. 

136.  Two  hills  each  264  feet  high  are  just  visible  from 
each  other  over  the  sea :  how  far  are  they  apart  ?     (Take 
the  radius  of  the  earth  =  4000  miles.)  Ans.  40  miles. 

137.  A   ship   sailing   out   of   harbor   is  watched  by  an 
observer  from  the  shore ;  and  at  the  instant  she  disappears 
below  the  horizon  he  ascends  to  a  height  of  20  feet,  and 
thus  keeps  her  in  sight  40  minutes  longer :  find  the  rate  at 
which  the  ship  is  sailing,  assuming  the  earth's  radius  to  be 
4000  miles,  and  neglecting  the  height  of  the  observer. 

Ans.  40V330  feet  per  minute. 

138.  From  the  top  of  the  mast  of  a  ship  64  feet  above 
the  level  of  the  sea  the  light  of  a  distant  lighthouse  is  just 
seen  in  the  horizon ;  and  after  the  ship  has  sailed  directly 
towards  the  light  for  30  minutes  it  is  seen  from  the  deck 
of  the  ship,  which  is  16  feet  above  the  sea :  find  the  rate 
at  which  the  ship  is  sailing.     (Take  radius  =  4000  miles.) 

Ans.  8V||  miles  per  hour. 


EXAMPLES.  203 

139.  A,  B,  C,  are  three  objects*  at  known  distances  apart ; 
namely,    AB  =  1056   yards,    AC  =  924  yards,    BC  =  1716 
yards.     An  observer  places  himself  at  a  station  P  from 
which  C  appears  directly  in  front  of  A,  and  observes  the 
angle  CPB  to  be  14°  24' :  find  the  distance  CP. 

Ana.  2109.824  yards. 

140.  A,  B,  C,  are  three  objects  such  that  AB  =  320  yards, 
AC  =  600  yards,  and  BC  =  435  yards.     From  a  station  P 
it  is  observed  that   APB  =  15°,  and  BPC  =  30° :   find  the 
distances  of  P  from  A,  B,  and  C ;  the  point  B  being  near- 
est to  P,  and  the  angle  APC  being  the  sum  of  the  angles 
APB  and  BPC.  Ana.  PA  =  777,  PB  =  502,  PC  =  790. 


204  PLANE  TRIGONOMETRY. 


CHAPTER  VIII. 

CONSTRUCTION  OF  LOGARITHMIC  AND  TRIGONOMETRIC 

TABLES, 

128.  Logarithmic  and  Trigonometric  Tables.  —  In  Chap- 
ters IV.,  V.,  and  VII.,  it  was  shown  how  to  use  logarithmic 
and  trigonometric  tables  ;  it  will  now  be  shown  how  to 
calculate   such   tables.     Although   the   trigonometric   func- 
tions are  seldom  capable  of   being   expressed  exactly,  yet 
they  can  be  found  approximately  for  any  angle  ;   and   the 
calculations  may  be  carried  to  any  assigned  degree  of  accu- 
racy.    We  shall  first  show  how  to   calculate   logarithmic 
tables,  and  shall  repeat  here  substantially  Arts.  208,  209, 
210,  from  the  College  Algebra. 

129.  Exponential  Series.  —  To  expand  ex  in  a  series  of 
ascending  powers  ofx. 

By  the  Binomial  Theorem, 


[2          »2 

nx(nx  —  \)(nx  —  2)  1_ 
~~          ~ 


12 

Similarly, 

i-i 


[2  |3 


(2) 


LOGARITHMIC  SERIES.  205 

and  therefore  series  (1)  is  equal  to  series  (2)  however  great 
n  may  be.  Hence  if  n  be  indefinitely  increased,  we  have 
from  (1)  and  (2) 


The  series  in  the  parenthesis  is  usually  denoted  by  e  ; 

hence  e*=  1  +  *  +  ^  +  ^  +  ^  +  .......     (3) 

|2     [3     [4 

which  is  the  expansion  of  ex  in  powers  of  x. 
This  result  is  called  the  Exponential  Theorem. 
If  we  put  x  =  1,  we  have  from  (3) 


From  this  series  we  may  readily  compute  the  approxi- 
mate value  of  e  to  any  required  degree  of  accuracy.  .This 
constant  value  e  is  called  the  Napierian  base  (Art.  64).  To 
ten  places  of  decimals  it  is  found  to  be  2.7182818284. 

Cor.  Let  a  =  ec  ;  then  c  =  loge  a,  and  a*  —  ecx.  Substi- 
tuting in  (3),  we  have 


or          a.=  l  +  *log.«  +  +...       (4) 

which  is  the  expansion  of  a*  in  powers  of  x. 

130.  Logarithmic   Series.  —  To  expand  loge(l  +  #)  in  a 
series  of  ascending  powers  of  x. 
By  the  Binomial  Theorem, 


-  1)  («-  2) 


206  PLANE   TRIGONOMETRY. 

=  1  +  X[a  _l  _  i  (a  ._  1)2+  i  .(a  _  1)3_  £(«_l)4+  ...] 
+  terms  involving  x2,  x",  etc. 

Comparing  this  value  of  ax  with  that  given  in  (4)  of  Art. 
129,  and  equating  the  coefficients  of  x,  we  have 

logea  =  a-l-^(a-l)2  +  i(a-l)3-i(^-l)4+- 
Put  a  =  1  +  #  ;  then 

log,(l+*)  =  *-f2  +  f-|4  +  .........     (3) 

This  is  the  Logarithmic  Series;  but  unless  x  be  very 
small,  the  terms  diminish  so  slowly  that  a  large  number  of 
them  will  have  to  be  taken  ;  and  hence  the  series  is  of  little 
practical  use  for  numerical  calculation.  If  x  >  1,  the  series 
is  altogether  unsuitable.  We  shall  therefore  deduce  some 
more  convenient  formulae. 

Changing  x  into  —  x,  (1)  becomes 

log,(l-»)  =  -*-f-f-f-  ......     (2) 

Subtracting  (2)  from  (1),  we  have 


Put 


1  —  x         n 
and  (3)  becomes 


i  !L±i=<>r  _  —      i         i       i 

°ge     n          "|_2w  +  1      3(2"  +  1)        5(2n+1) 
or  loge(n  -f  1) 


This  series  is  rapidly  convergent,  and  gives  the  logarithm 
of  either  of  two  consecutive  numbers  to  any  extent  when 
the  logarithm  of  the  other  number  is  known. 


COMPUTATION   OF  LOGARITHMS.  207 

131.  Computation  of  Logarithms.  —  Logarithms  to  the 
base  e  are  called  Napierian  Logarithms  (Art.  64).  They 
are  also  called  natural  logarithms,  because  they  are  the 
first  logarithms  which  occur  in  the  investigation  of  a 
method  of  calculating  logarithms.  Logarithms  to  the  base 
10  are  called  common  logarithms.  When  logarithms  are 
used  in  theoretical  investigations,  the  base  e  is  always 
understood,  just  as  in  all  practical  calculations  the  base 
10  is  invariably  employed.  It  is  only  necessary  to  com- 
pute the  logarithms  of  prime  numbers  from  the  series, 
since  the  logarithm  of  a  composite  number  may  be  obtained 
by  adding  together  the  logarithms  of  its  component  factors. 
The  logarithm  of  1  =  0.  Putting  n  =  1,  2,  4,  6,  etc.,  suc- 
cessively, in  (4)  of  Art.  130,  we  obtain  the  following 

Napierian  Logarithms  : 


log,  3  =log.2+2[J+  J-  +  J-+^  +  ..-]  =  1.09861228. 

log,  4  =  21oge2  =1.38629436. 

log.  5=lo 


log,  6  =  log,  2  +  log,  3  =1.79175946. 

log,  7  =  1og,6  +  2[i  +  §J^  +  ,-i§J+...]  =1.94590996. 

loge  8  =  31oge2  =2.07944154. 

loge  9  =  21oge3  =2.19722456. 

loge  10  =  loge  5  +  loge  2  =  2.30258509. 
And  so  on. 

The  number  of  terms  of  the  series  which  it  is  necessary 
to  include  diminishes  as  n  increases.     Thus,  in  computing 


208  PLANE  TRIGONOMETRY. 

the  logarithm  of   101,  the  first  term  of   the  series  gives 
the  result  true  to  seven  decimal  places. 

By  changing  b  to  10  and  a  to  e  in  (1)  of  Art.  65,  we  have 


or          common  logm  =  Napierian  logm  x  .43429448. 

The  number  .43429448  is  called  the  modulus  of  the  common 
system.  It  is  usually  denoted  by  /x. 

Hence,  the  common  logarithm  of  any  number  is  equal  to 
the  Napierian  logarithm  of  the  same  number  multiplied  by 
the  modulus  of  the  common  system,  .43429448. 

Multiplying  (4)  of  Art.  130  by  /x,  we  obtain  a  series  by 
which  common  logarithms  may  be  computed  ;  thus, 
loglo(n  +  l)  = 

(1) 

Common  Logarithms. 

Iog102  =  /ilog.2  =  .43429448  x  .69314718  =  .3010300. 
Iog103  =  /xloge3  =  .43429448  x  1.09861228  =  .4771213. 
Iog104  =  2  Iog102  =  .6020600. 

log,05  =  /xlog65  =  .43429448  x  1.60943790  =  .6989700. 
And  so  on. 

132.  If  6  be  the  Circular  Measure  of  an  Acute  Angle, 
sin  0,  0,  and  tan  0  are  in  Ascending  Order  of  Magnitude. 

With  centre  0,  and  any  radius,  de- 
scribe an  arc  BAB'.  Bisect  the  angle 
BOB'  by  OA;  join  BB',  and  draw  the 
tangents  BT,  B'T. 

Let  AOB  =  AOB'  =  0.     Then 

BB'  <  arc  BAB'  <  BT  +  B'T 

(Geom.,  Art.  246) 
.-.  BC  <  arc  BA  <  BT. 


LIMITING    VALUES  OF  SIN  0.  209 

EC      BA      BT 
'*  OB     OB      OB' 

.-.    sin  0  <  0  <  tan  0.  • 


133.  The  Limit  of  ^,  when  0  is  Indefinitely  Diminished, 
is  Unity. 

We  have       sin  0  <  0  <  tan  0  ......     (Art.  132) 

.-.  1<-A_<  sec0. 
sin0 

Now  as  0  is  diminished  indefinitely,  sec  6  approaches  the 
limit  unity  ;  then  when  9  =  0,  we  have  sec  0  =  1. 

A 

.-.  the  limit  of  -  —  ,  which  lies  between  sec0  and  unity, 
is  unity.  8in« 

.*.  also  -    -  approaches  the  limit  unity. 
0 


As        _  =  x  ge(}^  the  limifc  Q£  tan_0    when  $  .g 

60  6 

definitely  diminished,  is  also  unity. 
This  is  often  stated  briefly  thus  : 


=!,  and  =l,  when  0  =  0. 


NOTE.  —  From  this  it  follows  that  the  sines  and  the  tangents  of  very  small 
angles  are  proportional  to  the  angles  themselves. 

134.  If  9  is  the  Circular  Measure  of  an  Acute  Angle,  sin  0 

0s  B~ 

lies  between  9  and  9  --  ;  and  cos  9  lies  between  1  --  and 


(1)  We  have  tan->-      .....     (Art.  132) 

2i      2i 

06       0 

.,  sm-  >-  cos-. 


210  PLANE   TRIGONOMETRY. 


0        B  B 

.\  2 sin- cos r  >0cos2-- 

22  2i 

.-.  sin<9><9(l-sin2- 

a 


-"-}       .    .     (Art.  132) 
V        A/ 

.-.  sinO<0  and  >  0 - -• 


(2)  cos(9  =  l-2sin2-. 


Also,  .sin->--V-Yby  (1). 

22      4V2/ 


...  cose<l-2||-^ 


_     .     _      . 

2      16      512 


.-.  cos  B  >  1  -  T  and  <  1  -  -  +  —t. 
2  2      16 


NOTE.  —  It  may  be  proved  that  sin  6»>  0  —  — ,  as  follows : 

6 

We  have  3  sin  -  -  sin  9  =  4  sins  ?  ( Art.  50) (1) 

3  3 

.-.  3sin  -  -  sin  -=4B\n*  -  (by  putting  -  for  0) (2) 

3^  3  3^  3 


3  gin  .1  -  sin  —  =4  ei^  — (n) 

3n  3n-l  3n 

Multiply  (1),  (2),  ...  (n)  by  1,  3,  ...  3"-1,  respectively,  and  add  them, 

3n  Bin  1  _  8in  0  =  4(»in3  ?  +  3  sins  i  +  ...  S'1"1  sins  -^-V 
8"  \        3  32  3'1/ 


SINE  AND   COSINE  OF  10"  AND   OF  2'.  211 


^ 

~! /1  +  S+-^ri) (Art.  182) 


If  n  =  oo ,  then  — j—  =  1 (Art.  133) 

4      /        1  1^_4_^    _1 ? 

33  °A 1  +  3^  +  '"  g^2j  ~  "sT  '  1  _  ^  ~  7T 

32 

•    0-sin0<-,  and  /.  sin0>0--. 
6  6 

This  makes  the  limits  for  ein  0  closer  than  in  (1)  of  this  Art. 

135.  To  calculate  the  Sine  and  Cosine  of  10"  and  of  V. 

(1)  Let  0  be  the  circular  measure  of  10". 

Then 

g  =          1Q  *          =  3.141592653589793  •  •  - 
180  x  60  x  60  64800 

or         0  =  .000048481368110  •  •  -,  correct  to  15  decimal  places. 

.  -.  -  =  .000000000000032  •  • .,       «          «        «  « 

4 

...  0-~  =  . 000048481368078-.,       «          «         "  « 

4 

Hence  the  two  quantities  0  and  0 agree  to  12  deci- 

4 

03 

mal  places  ;  and  since  sin  6  <  0  and  >  0 (Art.  134), 

4 

.-.  sin  10"  =  .000048481368,  to  12  decimal  places. 
We  have 


cos  10"  =  VI -sin2 10"  =  1  -  i  sin2 10" 

=  .9999999988248  •••,  to  13  decimal  places. 

Or  we  may  use  the  results  established  in  (2)  of  Art.  134, 
and  obtain  the  same  value. 


212  PLANE  TRIGONOMETRY. 

(2)  Let  0  be  the  circular  measure  of  1'. 
Then 

0  = ^—  =  .000290888208665,  to  15  decimal  places. 

loU  X  oU 

.-.^=.000000000006        to  12         "  " 

4 

.-.  6  -  -  =  .00029088820          to  11         "  « 

4 

/j3 

Hence  0  and  0 differ  only  in  the  twelfth  decimal. 

4 

.-.  sin  1' =  .00029088820  to  11  decimal  places. 


cos  1'  =  Vl  -  sin2  1  '  =  .  999999957692025  to  15  decimal  places, 
Otherwise  thus  : 

1  _  61  =.999999957692025029  to  18  decimal  places. 

and      —  =  .00000000000000044  to  17  decimal  places. 
16 

Butcosl'>l-^and<l-f  +  ^  .     .     .     (Art.  134) 

2i  —         J-O 

.-.  cos  1'  =  .999999957692025  to  15  decimal  places,  as  before. 

Cor.  1.  The  sine  of  10"  equals  the  circular  measure  of  10", 
to  12  decimal  places  ;  and  the  sine  of  1'  equals  the  circular 
measure  of  V  to  11  decimal  places. 

Cor.  2.  Ifn  denote  any  number  of  seconds  less  than  60,  we 
shall  have  approximately 

sin  n"  =  n  sin  1", 

for  the  sine  of  n"  =  the  circular  measure  of  n",  approxi- 
mately, =  n  times  the  circular  measure  of  1". 


3    n  =  circular  measure  of  n"  oximately  .   that 

sml" 
is,  the   number  of   seconds  in  any  small  angle   is   found 


TABLES  OF  NATURAL  SINES  AND  COSINES. 

approximately  by  dividing  the   circular   measure  of  that 
angle  by  the  sine  of  one  second. 

136.  To  construct  a  Table  of  Natural  Sines  and  Cosines  at 
Intervals  of  1'. 

We  have,  by  Art.  45, 

sin(#  +  y}  =  2  sin  a; cosy  —  sin(o?  —  y), 
cos(#  +  y}  =  2  cos  a; cosy  —  cos  (a:  —  y). 

Suppose  the  angles  to  increase  by  1';  putting  y  —  1',  we 
have, 

sin(a  +  l')  =  2sina;cosl'-sin(a-l')  ....     (I) 

cos  (#  +  1')  =  2  cos  a;  cos  1'  —  cos(x  —  1')  .     ...     (2) 

Putting  x=  1',  2',  3',  4',  etc.,  in  (1)  and  (2),  we  get  for 
the  sines 

sin  2'  =  2  sin  1'  cos  1'  -  sin  0'  =  .0005817764, 

sin  3'  =  2  sin  2'  cos  1'  -  sin  V  =  .0008726646, 
sin  4'  =  2  sin  3'  cos  1'  -  sin  2'  =  .0011635526 ; 
and  for  the  cosines 

cos  2'  =  2  cos  1'  cos  V  -  cos  0'  =  .9999998308, 
cos  3'  =  2  cos  2'  cos  1'  -  cos  1'  =  .9999996193, 
cos  4'  =  2  cos  3'  cos  1'  -  cos  2'  =  .9999993223. 

We  can  proceed  in  this  manner*  until  we  find  the  values 
of  the  sines  and  cosines  of  all  angles  at  intervals  of  1'  from 
0°  to  30°. 

137.  Another  Method. 

Let  a  denote  any  angle.     Then,  in  the  identity, 

sin  (n  +  1)«  =  2  sin na  cos  a  —  sin  (n  —  l)a, 
put  2  (1  —  cos  a)  =  fc, 

and  we  get 

sin(n-j-l)«  —  sin  na=sinna  —  sin(w  — 1)«  —  k  sin  na  .     (1) 

*  This  method  ia  due  to  Thomas  Simpson,  an  English  geometrician. 


214  PLANE  TRIGONOMETRY. 

This  formula  enables  us  to  construct  a  table  of  sines  of 
angles  whose  common  difference  is  a. 

Tims,  suppose  a  =  10",  and  let  n  =  1,  2,  3,  4,  etc. 

Then 

sin  20"  -  sin  10"  =  sin  10"  -  k  sin  10", 

sin  30"  -  sin  20"  =  sin  20"  -  sin  10"  -  k  sin  20", 

sin  40"  -  sin  30"  =  sin  30"  -  sin  20"  -  k  sin  30",  etc. 

These  equations  give  in  succession  sin  20",  sin  30",  etc. 
It  will  be  seen  that  the  most  laborious  part  of  this  work  is 
the  multiplication  of  k  by  the  sines  of  10",  20",  etc.,  as  they 
are  successively  found.  But  from  the  value  of  cos  10",  we 
have 

A;  =  2  (1  -  cos  10")  =  .0000000023504, 

the  smallness  of  which  facilitates  the  process. 

In  the  same  manner  a  table  of  cosines  can  be  constructed 
by  means  of  the  formula, 

cos  (n  +  1)«  —  cos  na  —  cos  na  —  cos  (n  —  1) a  —  k  cos  ««, 
which  is  obtained  from  the  identity, 

cos  (71  +  1)«  =  2  cos  na  cos  a  —  cos  (n  —  l)a, 
by  putting  2(1  —  cos  a)  —  k,  as  before. 

138.  The  Sines  and  Cosines  from  30°  to  60°. —It  is  not 
necessary  to  calculate  in  this  way  the  sines  and  cosines  of 
angles  beyond  30°,  as  we  can  obtain  their  values  for  angles 
from  30°  to  60°  more  easily  by  means  of  the  formulae  (Art. 
45): 

sin  (30°  +  «)  =  cos  «  -  sin  (30°  -  a), 

cos  (30°  +  «)  =  cos  (30°  -  a)  -  sin  a, 
by  giving  a  all  values  up  to  30°.     Thus, 

sin  30°  V  =  cos  1'  -  sin  29°  59', 

cos  30°  V  =  cos  29°  59'  —  sin  1',  and  so  on. 


TABLES   OF  TANGENTS  AND   SECANTS.         215 

139.  Sines  of  Angles  greater  than  45°.  —  When  the  sines 
of  angles  up  to  45°  have  been  calculated,  those  of  angles 
between  45°  and  90°  may  be  deduced  by  the  formula 

sin  (45°  +  a)  -  sin  (45°  -  a)  =  V£  sin  a       .      (Art.  45) 

Also,  when  the  sines  of  angles  up  to  60°  have  been  found, 
the  remainder  up  to  90°  can  be  found  still  more  easily  from 
the  formula 

sin  (60°  +  «)  —  sin  (60°  —  «)  =  sin  «. 
Having  completed  a  table  of  sines,  the  cosines  are  known, 
Smce  cosa  =  sin(90°-<*). 

Otherwise  thus :  When  the  sines  and  cosines  of  the  angles 
up  to  45°  have  been  obtained,  those  of  angles  between  45° 
and  90°  are  obtained  from  the  fact  that  the  sine  of  an 
angle  is  equal  to  the  cosine  of  its  complement,  so  that  it  is 
not  necessary  to  proceed  in  the  calculation  beyond  45°. 

NOTE.  —  A  more  modern  method  of  calculating  the  sines  and  cosines  of  angles 
is  to  use  series  (3)  and  (4)  of  Art.  156. 

140.  Tables  of  Tangents  and  Secants.  —  To  form  a  table 
of  tangents,  we  find  the  tangents  of  angles  up  to  45°,  from 
the  tables  of  sines  and  cosines,  by  means  of  the  formula 

sin  a 

tan  a  = 

cos  a 

Then  the  tangents  of  angles  from  45°  to  90°  may  be 
obtained  by  means  of  the  identity  * 

tan  (45°  +  a)  =  tan  (45°  -  a)  +  2  tan  2  a. 

When  the  tangents  have  been  found,  the  cotangents  are 
known,  since  the  cotangent  of  any  angle  is  equal  to  the 
tangent  of  its  complement. 

A  table  of  cosecants  may  be  obtained  by  calculating  the 
reciprocals  of  the  sines;  or  they  may  be  obtained  more 

*  Called  Cagnoli's  formula. 


216  PLANE  TRIGONOMETRY. 

easily  from  the  tables  of  the  tangents  by  means  of  the 
formula 

cosec  a  =  tan  -  +  cot  a. 

Zi 

The  secants  are  then  known,  since  the  secant  of  any 
angle  is  equal  to  the  cosecant  of  its  complement. 

141.  Formulae  of  Verification.  —  Formulas  used  to  test  the 
accuracy  of  the  calculated  sines  or  cosines  of  angles  are  called 
Formula}  of  Verification. 

It  is  necessary  to  have  methods  of  verifying  from  time 
to  time  the  correctness  of  the  values  of  the  sines  and 
cosines  of  angles  calculated  by  the  preceding  method,  since 
any  error  made  in  obtaining  the  value  of  one  of  the  func- 
tions would  be  repeated  to  the  end  of  the  work.  For  this 
purpose  we  may  compare  the  value  of  the  sine  of  any  angle 
obtained  by  the  preceding  method  with  its  value  obtained 
independently. 

Thus,  for  example,  we  know  that  sin  18°  =  ^-"^-1  (Art. 

57)  ;  hence  the  sine  of  18°  may  be  calculated  to  any  degree 
of  approximation,  and  by  comparison  with  the  value  obtained 
in  the  tables,  we  can  judge  how  far  we  can  rely  upon  the 
tables. 

Similarly,  we  may  compare  our  results  for  the  angles 
221°,  30°,  36°,  45°,  etc.,  calculated  by  the  preceding  method 
with  the  sines  and  cosines  of  the  same  angles  as  obtained 
in  Arts.  26,  27,  56,  57,  58,  etc. 

There  are,  however,  certain  well-known  formulae  of  veri- 
fication which  can  be  used  to  verify  any  part  of  the  calcu- 
lated tables ;  these  are 

Euler^s  Formula3, : 
sin  (36°  +  A)  -  sin  (36°  -  A)  +  sin  (72°  -  A) 

-  sin  (72°  +  A)  =  sin  A. 
cos  (36°  +  A)  +  cos  (36°  -  A)  -  cos  (72°  +  A) 

-  cos  (72°  -  A)  =  cos  A. 


LOGARITHMIC  TRIGONOMETRIC  FUNCTIONS.    217 

Legendre's  Formula  : 
sin  (54°  +  A)  +  sin  (54°  -  A)  -  sin  (18°  +  A) 

-  sin  (18°  —  A)  =  cos  A. 

The  verification  consists  in  giving  to  A  any  value,  and 
taking  from  the  tables  the  sines  and  cosines  of  the  angles 
involved:  these  values  must  satisfy  the  above  equations. 

To  prove  Euler's  Formulas,  : 

sin  (36°  +  A)  -  sin  (36°  -  A)  =  2  cos  36°  sin  A    .     (Art.  45) 

V5  +  1 


/ 

sin  (72°  +  A)  -  sin  (72°  -  A)  =  2  cos  72°  sin  A    .     (Art.  45) 

=  ^5  ~  1  sin  A     .     (Art.  57) 

Subtracting  the  latter  from  the  former,  we  get  sin  A. 
Similarly,  Euler's  second  formula  may  be  proved. 
By  substituting  90°  —  A  for  A  in  this  formula  we  obtain 
Legendre's  Formula. 

142.  Tables  of  Logarithmic  Trigonometric  Functions.  - 

To  save  the  trouble  of  referring  twice  to  tables  —  first  to 
the  table  of  natural  functions  for  the  value  of  the  function, 
and  then  to  a  table  of  logarithms  for  the  logarithm  of  that 
function  —  it  is  convenient  to  calculate  the  logarithms  of 
trigonometric  functions,  and  arrange  them  in  tables,  called 
tables  of  logarithmic  sines,  cosines,  etc. 

When  tables  of  natural  sines  and  cosines  have  been  con- 
structed, tables  of  logarithmic  sines  and  cosines  may  ,be 
made  by  means  of  tables  of  ordinary  logarithms,  which  will 
give  the  logarithm  of  the  calculated  numerical  value  of  the 
sine  or  cosine  of  any  angle  ;  adding  10  to  the  logarithm  so 
found  we  have  the  corresponding  tabular  logarithm.     The 
logarithmic  tangents  may  be  found  by  the  relation 
log  tan  A  =  10  +  log  sin  A  —  log  cos  A  ; 
and  thus  a  table  of  logarithmic  tangents  may  be  constructed. 


218  PLANE  TRIGONOMETRY. 


PROPORTIONAL  PARTS. 

143.  The  Principle  of  Proportional  Parts.  —  It  is  often 
necessary  to  find  from  a  table  of  logarithms,  the  logarithm 
of  a  number  containing  more  digits  than  are  given  in  the 
table.     In  order  to  do  this,  we  assumed,  in  Chapter  IV.,  the 
principle  of  proportional  parts,  which  is  as  follows  : 

The  differences  between  three  numbers  are  proportional  to  the 
corresponding  differences  between  their  logarithms,  provided 
the  differences  between  the  numbers  are  small  compared  with 
the  numbers. 

By  means  of  this  principle,  we  are  enabled  to  use  tables 
of  a  more  moderate  size  than  would  otherwise  be  necessary. 

We  shall  now  investigate  how  far,  and  with  what  excep- 
tions, the  principle  or  rule  of  proportional  increase  is  true. 

144.  To  prove  the  Rule  for  the  Table  of  Common  Loga- 
rithms. 

We  have 


where  /*  =  .43429448  •••,  a  quantity  <  j. 

Now  let  n  be  an  integer  not  <  10000,  and  d  not   >  1  ; 

then  -  is  not  greater  than  .0001. 

...  j£#  is  not  >^(.0001)2,  i.e.,  not  >  .0000000025  ; 

and        &—  is  much  less  than  this. 
3n3 

.-.  log  (n  -f-  d)  —  log  n  —  //,-,  correct  at  least  as  far  as  seven 
decimal  places. 


RULE  OF  PROPORTIONAL  PARTS.  219 

Hence  if  the   number  be  changed  from  n  to  n  -f  d,  the 
corresponding   change   in  the  logarithm  is  approximately 
£f 
n 

Therefore,  the  change  of  the  logarithm  is  approximately 
proportional  to  the  change  of  the  number. 

145.  To  prove  the  Rule  for  the  Table  of  Natural  Sines. 

sin  (0  -\-  h)  —  sin  6  —  sin  h  cos  0  —  sin  6  (1  —  cos  h) 

=  sin  h  cos  6  (l  -  tan  0  tan  -\  .    (Art.  51) 

\  4/ 

If  h  is  the  circular  measure  of  a  very  small  angle,  sin  h  =  h 

nearly,  and  tan  ^  =  ^  nearly. 

2      2 

.-.  sin  (0  +  h)  -  sin  0  =  7i  cos  0 fl  -  tan  0  tan  - 

\ 

7,2 

=  7i  cos  e  —  ~  sin  0. 

2i 

If  7i  is  the  circular  measure  of  an  angle  not  >  1',  then 

7i  is  not  >  .0003  (Art.  135).  .-.  -  is  not  >  .00000005;  and 
sin  6  is  not  >  1. 

.-.  sin(0-t-7i)  —sin  6=hcos  0,  as  far  as  seven  decimal  places, 
which  proves  the  proposition. 

Similarly,  sin  (0  —  h)  —  sin  6  =  —  h  cos  0,  approximately. 

146.  To  prove  the  Rule  for  a  Table  of  Natural  Cosines. 

cos  (0  —  h)  —  cos  0  =  sin  h  sin  0  —  cos  0(1—  cos  h) 

=  sin  h  sin  0(1  —  cot  0  tan  -  )• 


V  2, 

If  h  is  the  circular  measure  of  a  very  small  angle,  sin  h  =  h 

nearly,  and  tan  -  =  -  nearly. 

.-.  cos  (0  —  h)  —  cos  0  =  h  sin  0/1  —  cot  0  tan  - 

-COS0. 


220  PLANE  TRIGONOMETRY. 

We  may  prove,  as  in  Art.  145,  that 

—  cos  0  is  not  >  .00000005. 

.*.  cos  (0  —  h)  —  cos  0  =  h  sin  0,   as   far   as  seven  decimal 
places,  which  proves  the  proposition. 

Similarly,  cos  (6  +  h)  —  cos  0  =  —  h  sin  0,  approximately. 

147.  To  prove  the  Rule  for  a  Table  of  Natural  Tangents. 
sin  sin* 


cos  (6  +  h)      cos  0     cos  (0  +  7i)  cos  0 

_  tan  h 

~cos20(l-tan0tan7i)' 

If  h   is   the   circular   measure   of   a   very   small   angle, 
tan  7i  =  h  nearly. 

.-.  tan  (0  +  7i)  -  tan  0  =  -I1  sec2  ° 


l-h  tan  0 


.-.  tan  (6  -f  h)  —  tan  6  =  h  sec2  0,  approximately, 
unless  sin  0  sec3  6  is  large,  which  proves  the  proposition. 
Similarly,  cot  (0  —  h)  —  cot  0  =  7i  cosec2  0,  approximately. 

/#c7i.  1.  If  7i  is  the  circular  measure  of  an  angle  not  >  lr, 
then  h  is  not  >  .0003.  Hence  the  greatest  value  of  h2  sin  0 
sec3  6  is  not  >  .00000009  sin  0  sec3  0.  Therefore,  when  0  >  |, 

we  are  liable  to  an  error  in  the  seventh  place  of  decimals. 
Hence  the  rule  is  not  true  for  tables  of  tangents  calculated 
for  every  minute,  when  the  angle  is  between  45°  and  90°. 

Sch.  2.  Since  the  cotangent  of  an  angle  is  equal  to  the 
tangent  of  its  complement,  it  follows  immediately  that  the 
rule  must  not  be  used  for  a  table  of  cotangents,  calculated 
for  every  minute,  when  the  angle  lies  between  0°  and  45°. 


RULE  OF  PROPORTIONAL  PARTS.  221 

148.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Sines. 

sin  (0  +  h)  —  smO  =  h  cos  0  -  -  sin  0     .     .     (Art. 


. 

sin  (9 

.-.  log  sin  (0  +  h)  —  log  sin  0 


=  Jh  cot  (9  -  h*  --(h  cot  (9  -  7|Y+  -.."I     (Art.  130) 
L  2      2\  2J          J 

=  ^  cot  0  -        (1  +  cot2  0)  +  ... 


If  ft  is  the  circular  measure  of  an  angle  not  >  10",  then 
h  is  not  >  .00005,  and  therefore,  unless  cot  6  is  small  or 
cosec2  0  large,  we  have 

log  sin  (0  +  h)  —  log  sin  0  =  pli  cot  0, 

as  far  as  seven  decimal  places,  which  proves  the  rule  to  be 
generally  true. 

JSch.  1.  When  0  is  small,  cosec  9  is  large.  If  the  leg 
sines  are  calculated  to  every  10",  then  h  is  not  >.  00005, 
and  /A  is  not  >  .5. 

cosec2  0  is  not 


In  order  that  this  error  may  not  affect  the  seventh  decimal 
place,  6  cosec2  0  must  not  be  >  103,  that  is,  0  must  not  be 
less  than  about  5°. 

When  6  is  small,  cot  0  is  large.     Hence,  when  the  angles 


222  PLANE  TRIGONOMETRY. 

are  small,  the  differences  of  consecutive  log  sines  are  irregu- 
lc$,  and  they  are  not  insensible.  Therefore  the  rule  does 
not  apply  to  the  log  sine  when  the  angle  is  less  than  5°. 


.  2.    When  0  is  nearly  a  right  angle,  cot#  is  small, 
and  cosec  0  approaches  unity. 

Hence,  when  the  angles  are  nearly  right  angles,  the  dif- 
ferences of  consecutive  log  sines  are  irregular  and  nearly 
insensible. 

149.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Co- 
sines. 

cos  (6  -  h)  -  cos  0  =  h  sin  0  -     cos  0  .     (Art.  146) 


cos  0 
.-.  log  cos  (0  —  h)  —  log  cos  0 


=  /*  [ 


h  tan  0  -  —  -  1  fh  tan  0  - 


In  this  case  the  differences  will  be  irregular  and  large 
when  0  is  nearly  a  right  angle,  and  irregular  and  insensible 
when  0  is  nearly  zero.  This  is  also  clear  because  the  sine 
of  an  angle  is  the  cosine  of  its  complement. 

150.  To  prove  the  Rule  for  a  Table  of  Logarithmic  Tan- 
gents. 

tan  (6  +  h)  —  tan  0  =  h  sec2  0  +  7t2  sin  0  sec3  0  .     (Art.  147) 


tan  &  sin  $  cos  0 


RULE  OF  PROPORTIONAL  PARTS.  223 

.-.  log  tan  (0  +  h)  —  log  tan  6 


.-.  log  tan  (0  +  h)  —  log  tan  0 

ph        _2    *  2  cos  20 
~  sin  0  cos  0        **    sin2  20 

151.  Cases  where  the  Principle  of  Proportional  Parts  is 
Inapplicable. 

It  appears  from  the  last  six  Articles  that  if  h  is  small 
enough,  the  differences  are  proportional  to  h,  for  values  of  0 
which  are  neither  very  small  nor  nearly  equal  to  a  right  angle. 

The  following  exceptional  cases  arise  : 

(1)  The  difference  sin  (0  -f-  h)  —  sin  6  is  insensible  when 
0  is  nearly  90°,  for  in  that  case  hcosO  is  very  small;  it  is 
then  also  irregular,  for  Jft2sin0  may  become  comparable 
with  ft  cos  0. 

(2)  The  difference  cos  (0  -f  h)  —  cos  6  is  both  insensible 
and  irregular  when  6  is  small. 

(3)  The  difference  tan  (0  +  h)  —  tan  0  is  irregular  when 
0  is  nearly  90°,  for  ft2  sin  0  sec3  0  may  then  become  compar- 
able with  h  sec2  0  ;  it  is  never  insensible,  since  sec  0  is  not 
<1. 

(4)  The  difference  log  sin  (0  -f-  ft)  —  log  sin  0  is  irregular 
when  6  is  small,  and  both  irregular  and  insensible  when  0  is 
nearfy  90°. 

(5)  The  difference  log  cos  (6  +  ft)  —  log  cos  6  is  insensible 
and  irregular  when   0  is  small,  and  irregular  when   0   is 

90°. 


(6)  The  difference  log  tan  (0  +  7i)  —  log  tan  0  is  irregular 
when  0  is  either  small  or  nearly  90°. 


224  PLANE   TRIGONOMETRY. 

A  difference  which  is  insensible  is  also  irregular;  but  the 
converse  does  not  hold. 

When  the  differences  for  a  function  are  insensible  to  the 
number  of  decimal  places  of  the  tables,  the  tables  will  give 
the  functions  when  the  angle  is  known,  but  we  cannot  use 
the  tables  to  find  any  intermediate  angle  by  means  of  this 
function;  thus,  we  cannot  determine  0  from  the  value  log 
cos  0,  for  small  angles,  or  from  the  value  log  sin  &,  for  angles 
nearly  90°. 

When  the  differences  for  a  function  are  irregular  without 
being  insensible,  the  approximate  method  of  proportional 
parts  is  not  sufficient  for  the  determination  of  the  angle  by 
means  of  the  function,  nor  the  function  by  means  of  the 
angle  ;  thus,  the  approximation  is  inadmissible  for  log  sin  0, 
when  6  is  small,  for  log  cos  6,  when  0  is  nearly  90°,  and 
for  log  tan  6  in  either  case.  (Compare  Art.  81.) 

In  these  cases  of  irregularity  without  insensibility,  the 
following  three  means  may  be  used  to  effect  the  purpose  of 
finding  the  angle  corresponding  to  a  given  value  of  the 
function,  or  of  the  function  corresponding  to  a  given  angle.* 

152.  Three  Methods  to  replace  the  Rule  of  Proportional 
Parts. 

(1)  The  simplest  plan  is  to  have  tables  of  log  sines  and 
log  tangents,  for  each  second,  for  the  first  few  degrees  of 
the  quadrant,  and  of  log  cosines  and   log  cotangents,  for 
each  second,  for  the  few  degrees  near  90°.     Such  tables  are 
generally  given  in  trigonometric  tables  of   seven  places ; 
we  can  then  use  the  principle  of  proportional  parts  for  all 
angles  which  are  not  extremely  near  0°  or  90°. 

(2)  Delambre's  Method.    In  this  method  a  table  is  con- 
structed which  gives  the  value  of  log  — — (-  log  sin  I"  for 

9 

every  second  for  the  first  few  degrees  of  the  quadrant. 

*  This  article  has  been  taken  substantially  from  Hobson's  Trigonometry. 


METHODS   TO  REPLACE   THE  RULE.  225 

Let  6  be  the  circular  measure  of  n  seconds.  Then,  when 
0  is  small,  we  have  6  =  n  sin  1",  approximately. 

T      sin0      T       sinn"       ,        •      „      ^ 
.-.  log  — —  =  log — : — —  =  log  sin  n"  —  log  n  —  log  sm  1". 
0  n  sin  1 

.-.  log  sin  n"  =  log  n  +  ( log— — h  log  sin  1"  Y 
V         0.  / 

Hence,  if  the  angle  is  known,  the  table  gives  the  value 
of  the  expression  in  parenthesis,  and  log  ft  can  be  found 
from  the  ordinary  table  of  the  logs  of  numbers ;  thus  log 
sin  n"  can  be  found. 

If  log  sin  n"  is  given,  we  can  find  approximately  the  value 
of  n,  and  then  from  the  table  we  have  the  value  of  the 
expression  in  parenthesis;  thus  we  can  find  logw,  and  then 
n  from  an  ordinary  table  of  logs  of  numbers. 

Rem.   When  0  is  small  (less  than  5°), 

=  1  —  — ,  approximately      .     .     (Art.  134,  Note) 

(7  6 

Hence  a  small  error  in  0  will  not  produce  a  sensible  error 

in  the  result,  since  log  -    -  will  vary  much  less  rapidly 
than  6. 

(3)  Maskelyne' s  Method.  The  principle  of  this  method  is 
the  same  as  that  of  Delambre's.  If  0  is  a  small  angle,  we 
have 

/\5 

sin  0  =  0 ,  approximately, 

6 

and  cos  0=1--,  "  .     .     (Art.  134) 

sinfl 
"     0 


.-.  log  sin  6=  log  0  +  |  log  cos  0,  approximately. 


226  PLANE  TRIGONOMETRY. 

When  6  is  a  small  angle,  the  differences  of  log  cos  6  are 
insensible  (Art.  149)  ;  hence,  if  0  be  given,  we  can  find 
log  0  accurately  from  the  table  of  natural  logarithms,  and 
also  an  approximate  value  of  log  cos  0  ;  the  formula  then 
gives  log  sin  0  at  once. 

If  log  sin  0  be  given,  we  must  first  find  an  approximate 
value  of  0  from  the  table,  and  use  that  for  finding  log  cos  0, 
approximately  ;  6,  is  then  obtained  from  the  formula. 

EXAMPLES. 

1.   Prove  l      2.    +  i+...  =  2e. 


q        1         /        o  q 

.   Prove  log-  =  --[  —  -  -  +  -  - 

b2      2       1-3'23      2-5 


5.   Prove  tan  0  +  \  tan3  0  -f  |  tan5  0  H 


6.  Prove  loge  11  =  2.39789527  .••>  by  (4)  of  Art.  130. 

7.  Prove  loge  13  =  2.56494935".,         «  " 

8.  Prove  loge  17  =  2.83321334..-,         "  " 

9.  Prove  log.  19  =  2.  9444394   •••,         «  " 

10.  Find,  by  means  of  the  table  of  common  logarithms 
and  the  modulus,  the  Napierian  logarithms  of  1325.07, 
52.9381,  and  .085623.  Ana.  7.18923,  3.96913,  -  2.4578. 


EXAMPLES.  227 


/Q 

11.  Prove  that  the  limit  of  m  sin  —  is  0,        when  ra  =  oo. 

m 

12.  "  "  "         ratan  —  is  0, 

m 

13.  "  "  "         —  sin  —  is  Trr2,      "      n  =  oc. 

2          n 

14.  "  "  "         Trr2  tan  _  is  Trr2,        "      n  =  oo. 

n 


15.  "  "  "  -  is  — ,  "      0  =  0. 

vers  60       o2 

16.  "  "  "         fcos^Yisl,  "      n  =  oo. 

/  ff\r 

17.  "  a  "         [sin-]isl,  "      n  =  oo. 

V    »y 

18.  "  "  "         /^cos-Yisl,  "      n  =  oo. 

V       »/ 

0  n 

/sin  _N 

19.  "  I 5]isl,  "      ?i  =  oo. 

/  /}\  ^i^  ^^ 

20.  «  "  "         (cos-j    ise~^,       "      n  =  oo. 

21.  "  "  "         (cos-)    is  zero,  when  n  =  oo. 


22.    If  0  is  the  circular  measure  of  an  acute  angle,  prove 
(1)  cos0<l-     +     ,and  (2) 


23.   Given  =  1013.  prOve  that  0  =  4°  24',  nearly. 

6         1014 


24.    Given  —  =          :  prove  that  0  =  3°,  nearly. 
0        2166 


228  PLANE   TRIGONOMETRY. 

25.  Given  sin  <£  =  n  sin  6,  tan  <f>  =  2  tan  0  :  find  the  limit- 
ing values  of  n  that  these  equations  may  coexist. 

Ans.  n  must  lie  between  1  and  2,  or  between  —  1  and  —  2. 

26.  Find  the  limit  of  (cos  ax)cosec26x,  when  x  =  0. 


27.  From  a  table  of  natural  tangents  of  seven  decimal 
places,  show  that  when  an  angle  is  near  60°  it  may  be 
determined  within  about  -  of  a  second. 


28.  When  an  angle  is  very  near  64°  36',  show  that  the 
angle  can  be  determined  from  its  log  sine  within  about  y1^ 
of  a  second  ;  having  given  (loge  10)  tan  64°  36'  =  4.8492,  and 
the  tables  reading  to  seven  decimal  places. 


DE  MOIVRE'S   THEOREM.  229 


CHAPTER   IX. 
DE  MOIVKE'S   THEOREM,*  —  APPLICATIONS, 

153.   De  Moivre's  Theorem.  —  For  any  value  of  n,  positive 
or  negative,  integral  or  fractional. 


.     (1) 

I.    When  n  is  a  positive  integer. 

We  have  the  product 
(cos  a  +  V—  1  sin  a)  (cos  ft  +  V—  1  sin  J3) 

=  (cos  a  cos  (3  —  sin  a  sin  ft)  -f  V  —  1  (cos  a  sin  ft  +  sin  a  cos  /?) 

=  cos  (a  -f  £)  +  V^T  sin  (a  +  /?)  . 

Similarly,  the  product 
[cos  («  +  £)  +  V11!  sin  (a  +  0)  ]  [cos  y  +  V^~l  sin  7] 

=  cos  (a  4-  /?  +  y)  +  V^^  sin  (a  +  ft  -f  y)  - 

Proceeding  in  this  way  we  find  that  the  product  of  any 
number  n  of  factors,  each  of  the  form 

cosa-{-  V—  1  sin  a  =  cos  (a  +  /?  +  y  H  ----  w  terms) 
-f  V^T  sin  (a  +  /?  +  y  -|  ----  n  terms). 

Suppose  now  that  a  =  /?  =  y  =  etc.  =  0,  then  we  have 

(cos  0  +  V^T  sin  0)H  =  cos  n6  +  V^l  sin  w0, 
which  proves  the  theorem  when  n  is  a  positive  integer. 

*  From  the  name  of  the  French  geometer  who  discovered  it. 


230  PLANE  TRIGONOMETRY. 

II.    Wlien  n  is  a  negative  integer. 
Let  n  =  —  m;  then  m  is  a  positive  integer.     Then 
(cos  0  +  V^  sin  6)  *  =  (cos  0  +  V^T  sin  0)  ~m 
1  1 


-(by  I.) 


(cos0-f  V—  1  sili  0)m      cosra0-f-  V—  1  sinm0 

1 cosmfl  — V  —  1  sinmfl 

cosm^  -|-  V—  1  sin  mO      cosmO  —  V—  1  sinw0 

cosm^  —  V—  1  sinm0  / — ^   •       f\ 

—  —  —  =  cosm0  —  V—  Ismm^ 

cosj  mv  +  smj  mB 


=  cos  (—  mO)  +  V—  1  sin  (—mO). 
.  •.  (cos  6  4-  V^^  sin  6)  n  =  cos  nB  +  V^  sin  n0, 
which  proves  the  theorem  when  n  is  a  negative  integer. 

III.    Wlien  n  is  a  fraction,  positive  or  negative. 
Let  n  =  -,  where  p  and  q  are  integers.     Then 


(cos0+V:::Ism0)?'=cosp0+  V^sinp^by  I.  and  II.). 

/       P  P  \q 

But  f  cos  -0+  V  —  1  sin-0  j  —  cospO  +  V— 


.-.   (cos  0  +  V^l  sin  0)p  =    cos  -0  +  V^I  sin  -  6> 

- 

that  is,  one  of  the  values  of  (cos  0  +  V—  1  sin  6)q 

pO        -  .    p$ 

is  cos  --  hv—  Ism  — 

In  like  manner, 

(cos  0  —  V—  1  sin  0)n  =  cos  nO  —  V—  1  sin  nO. 

Thus,  De  Moivre's  Theorem  is  completely  established. 
'It  shows  that  to  raise  the  binomial  cos#  +  V—  1  sin^  to 


DE  MOIVRE'S   THEOREM.  231 

any  power,  we  have  only  to  multiply  the  arc  6  by  the 
exponent  of  the  power.  This  theorem  is  a  fundamental 
one  in  Analytic  Mathematics. 


154.  To  find  All  the  Values  of  (cos0  +  V^  sin  0)  — 
When  n  is  an  integer,  the  expression  (cos  0  +  V—  1  sin0)n 

P 
can  have  only  one  value.     But  if  n  is  a  fraction  =  -,  the 

expression  becomes 

_  P        ,  -  —  —  —  - 

(cos  6  +  V—  1  sin  0)*  =\  (cos  0  +  V  —  1  sin  0)p, 

which  has  q  different  values,  from  the  principle  of  Algebra 

(Art.  235).     In  III.  of  Art.  153,  we  found  one  of  the  values 

_  P 

of  (cos  6  -\-  V—  1  sin0)7;  we  shall  now  find  an  expression 

p 
which  will  give  all  the  q  values  of  (cos  9  -f  V—  1  sin  0)*. 

Now  both  cos  0  and  sin  0  remain  unchanged  when  0  is 
increased  by  any  multiple  of  27r;  that  is,  the  expression 
cos0-h  V—  Isin0  is  unaltered  if  for  0  we  put  (0-f  2r?r), 
where  r  is  an  integer  (Art.  36)  . 


=  [cos  (0  +  2rv)  +  V-  1  sin  (0  +  2r7r)]? 

^^  (Art.  153)  (1) 
</  «/' 

The  second  member  of  (1)  has  q  different  values,  and  no 
more;  these  q  values  are  found  by  putting  ?*=0, 1,  2,  •••  q— 1, 
successively,  by  which  we  obtain  the  following  series  of 
angles. 

p(0+2r7r)  pO 

When  r  =  0,       cos =  cos  — 


a         r  __  2  " 

etc.  etc. 


232  PLANE   TRIGONOMETRY. 

p(0  +  2nr) 
When  r=q—  1.  cos  ~  —         --  =  cos 


q 

J)(0  +  2gi 

=  cos-        — 

All  these  q  values  are  different. 

j>(0  +  2nr) 
When  ?'  =  g,  cos  -  =  cos 


fpO  \  pO 

=  COS  f          +  2p7TJ=  COS  y, 


the  same  value  as  when  r  =  0. 
When  r  =  q  +  1,  c< 


the  same  as  when  r  =  1, 

etc., 

from  which  it  appears  that  there  are  q  and  only  q  different 
values  of  cos^  —        —  ,  since  the  same  values  afterwards 


recur  in  the  same  order. 


Similarly  for  sin  — 
Therefore  the  expression 


cos 


gives  all  the  q  values  of  (cos  0  +  V—  1  sin  0) *  and  no  more. 
And  this  agrees  with  the  Theory  of  Equations  that  there 
must  be  q  values  of  a?,  and  no  more,  which  satisfy  the  equa- 
tion xq  =  c,  where  c  is  either  real  or  of  the  form  a-f-6  V— 1. 


APPLICATIONS   OF  DE  MOIVRE'S   THEOREM.     233 
APPLICATIONS  OF  DE  MOIVRE'S  THEOREM. 

155.  To  develop  cos  nO  and  sin  nO  in  Powers  of  sin  0  and 

COS0. 

We  shall  generally  in  this  chapter  write  i  for  V  —  1  in 
accordance  with  the  usual  notation.  * 

By  De  Moivre's  Theorem  (Art.  153)  we  have 

cos  ?i0H-  i  sinrj0  =  (cos0-Msin0)n  .     .     .     .     .     (1) 

Let  n  be  a  positive  integer.  Expand  the  second  member 
of  (1)  by  the  binomial  theorem,  remembering  that  i2=—  1, 
p  =  —  i}  and  that  i4  =  -fl  (Algebra,  Art.  219).  Equate 
the  real  and  imaginary  parts  of  the  two  members.  Thus, 

cos  nO  =  cosn  B  -  n(n~1)  cos"-2  0  sin2  0 

L? 

+  n^~ 


II 


n(n  -  *)  (n  ~  2) 


sin  n0  =  n  cos"-T0  sin  0  -         -  ~      cos*-3<9  sin80 

+  n(n-l)  (n-2)  (n-3)  (n  -  4)  COB.^sin5g_etc>   (3) 

[^ 

The  last  terms  in  the  series  for  cos  nO  and  for  sin  nO  will 
be  different  according  as  n  is  even  or  odd. 

The  last  term  in  the  expansion  of  (cos  0  +  isin0)n  is 
?'"sinn0;  and  the  last  term  but  one  is  nin  '^cos  0  sin""1^. 
Therefore  : 

When  n  is  even,  the  last   term  of  cosw0  is  t*sin*0  or 

n 

(  —  I)"1  sinw  0,  and  the  last  term  of  sin  nO  is  wn~-  cos  6  sin"-1  0 

or  n(—  1)  2    cos  0  sin""1  6. 

When  n  is  odd,  the  last  term  of  cos  nO  is  nin~l  cos  0  sin""1  9 

n—\ 

or  n(—  1)  2    cos  0  sin""1  0,  and  the  last   term   of  sinnO  is 

n—l 
n-l  »  2  n 


234  PLANE   TRIGONOMETRY. 

EXAMPLES. 

Prove  the  following  statements  : 

1.  sin 40  =  4  cos3  0  sin  0  —  4  cos 0  sin3 0. 

2.  cos%  0  =  cos4  0-6  cos2  0  sin2  0  +  sin4  0. 

156.  To  develop  sin  0  and  cos  0  in  Series  of  Powers  of  0. 

Put  nO  =  a  in  (2)  and  (3)  of  Art.  155;  and  let  n  be  in- 
creased without  limit  while  a  remains  unchanged.     Then 

since  0  =  -,  0  must  diminish  without  limit.     Therefore  the 

n 
above  formulas  may  be  written 


na      «(«  —  0)       n  9  /,  /sin  0V 

=  cos"0  --  ^—  —  ^-cosn~20(  ) 

[2  \    0    J 

<*(<*-  0)0*  -20)0*  -30)  cog 


4  ,  /sinL0V_ 

\  e  ) 


and  Bina^ 


«(«-fl)(«-2fl)          30/sinfly     .  ,2v 

[3  V  6»  y 

If  n  =  oo,  then  0  =  0,  and  the  limit  of  cos  0  and  its  powers 
is  1;  also  the  limit  of  (  S^11— )  and  its  powers  is  1.  Hence 
(1)  and  (2)  become  > 

cos«=l-|2  +  |4-|'+- (3) 


«3  ,   a5 
sm  «  =  «  —  —  +  — 


i.    In  the  series  for  since  and  cos  «,  just  found,  a  is  the 
circular  measure  of  the  angle  considered. 


CONVERGENCE  OF  THE  SERIES.  235 

Cor.  1.  If  a  be  an  angle  so  small  that  a2  and  higher 
powers  of  a  may  be  neglected  when  compared  with  unity, 
(3)  becomes  cos  a  =  1,  and  (4),  sin  a  =  a. 

If  a2,  «3  be  retained,  but  higher  powers  of  a  be  neglected, 
(3)  and  (4)  give 

3  2 

sin  «  =  a  —  —  ;  cos  a  =  1  —  —    (Compare  Art.  134) 


Cor.  2.   By  dividing  (3)  by  (4),  we  obtain 


O 


,  ,, 

O  •  O          O  •  O  •  O  •  4 


etc.      .     .     (5) 


157.  Convergence  of  the  Series.  —  The  series  (3)  and  (4) 
of  Art.  156  may  be  proved  to  be  convergent,  as  follows  : 

The  numerical  value  of  the  ratio  of  the  successive  pairs 
of  consecutive  terms  in  the  series  for  sin  «  are 


etc. 


2-3    4-5     6-7     8-9 


Hence  the  ratio  of  the   (n-fl)th  term  to  the  nth  term  is 

a2 

;  and  whatever  be  the  value  of  a.  we  can  take 

' 


n  so  large  that  for  such  value  of  n  and  all  greater  values, 
this  fraction  can  be  made  less  than  any  assignable  quantity  ; 
hence  the  series  is  convergent. 

Similarly,  it  may  be  shown  that  the  series  for  cos  a  is 
always  convergent. 

158.  Expansion  of  cosn0  in  Terms  of  Cosines  of  Multi- 
ples of  0,  when  n  is  a  Positive  Integer. 

Let     x  =  cos  0  +  i  sin  0  ; 

then       -  =  -  =  cos  0  —  i  sin  0. 
x     cos  0  -f  i  sin  0 

and  x  —  -  =  2isiuO    .....     (1) 

•          X 


236  PLANE  TRIGONOMETRY. 

Also  xn=  (cos  6+i  sin  0)"  =  cos  nO+i  sin  nO  (Art.  153)  (2) 
and        —  =  (cos0  —  isin0)M  =  cos?i0  —  i'sinntf  ...     (3) 

M? 

^.  2  cos  nO  =  xn  +  -i,  and  2  i  sin  n0  =  zn  -  i  .     (4) 

3J  3JM 

Hence  (2cos0)«  =  (#  +  or1)",  by  (1), 

=  xn  -f  nof-2  +  n(n~  JJ.  of-4  _j_  etc.  +  war<M-2)  +  orn 

I? 

IV  nor^  +  ^-\  +  ^=^lx^  +    1  V  etc. 
a  !  2  ~ 


2  cos  (»i  -  2)6  +        ~      2cos(yi  -  4)0  +  etc. 
l£ 

.-.  2"-1  cosn  ^  =  cos  nO  +  ?i  cos  (w  —  2)  0 

+  7ll!^ll)  Cos  O  -  4)  0  +  etc.        (5) 

NOTE.—  In  the  expansion  of  (x  +  x~l)n  there  are  w  +  1  terms;  thus  when  n  is 
even  there  is  a  middle  term,  the  (-  +  l)th,  which  is  independent  of  0,  and  which  is 


Hence  when  n  is  even  the  last  term  in  the  expansion  of  2n~l  cosn  6  is 


When  n  is  odd  the  last  term  in  the  expansion  of  2n—1  cosn  0"is 

159.   Expansion  of  sin"  9  in  Terms  of  Cosines  of  Multiples 
of  0,  when  n  is  an  Even  Positive  Integer. 

(2  %  sin  BY  =  (x-  *Yby  (1)  of  Art.  158 
V       XJ 

iT~ 

.^(n-1 

[2 


EXPANSION   OF  SINn0. 


237 


I? 


=  cos  n<9  -  w  cos  (n  —  2)0  +  n(n~1)  cos  (w  -  4)  ) 


4- 


160.  Expansion  of  sinrt  0  in  Terms  of  Sines  of  Multiples  of 
0,  when  n  is  an  Odd  Positive  Integer. 

(2  f  sin  ^)w  =  fx  -  -Yby  (1)  of  Art.  158 

V      XJ 


[2 


[ 


(2esin0)n 
=  2  1  sin  w^  —  n2t  sin  (n  —  2)0 


l± 


liGLzi 


of  Art  158] 


238  PLANE  TRIGONOMETRY. 

Whence  dividing  by  2i,  we  have 


H-l 

2  sin"0 


=  sin  nB  -  n  sin  (n  -2)0-}-  nn~      sin  (w  _  4)$ 

14 

(-  1)^(.  -l)...i(.  +  8) 


EXAMPLES. 

Prove  that 

1.  128  cos80=cos  8  0+8  cos  60  +  28  cos  4  0+56  cos  20+35. 

2.  64  cos70  =  cos  7  0  +  7  cos  50  +  21  cos  30  +  35  cos  0. 

161.  Exponential  Values  of  Sine  and  Cosine. 

Since     e  =  1  +  x  +     +     +      +  .-•     .     .     .     (Art.  129) 


=  cos  0  +  i  sin  0    .     .     .....     (Art.  156) 

/J2         /)4 

and        e-«  =  1_     +     _et, 


=  cos  0  —  i  sin  0. 
.-.  2  cos  0  =  eie  +  e-*0,  and  2  i  sin  0  =  e*9  —  e~i9      .     .     (1) 

e»0  _l_  e-tfl  .  e»»  _  e-f» 

.-.  cos0  =  —         —  ,  and  sin  0  =  -  :  —      ...     (2) 

—  2i  l 

which  are  called  the  exponential  values*  of  the  cosine  and 
sine. 

Cor.   From   these   exponential   values    we    may   deduce 
similar  values  for  the  other  trigonometric  functions.    Thus, 


(3) 


*  Called  also  Euler's  equations,  after  Euler,  their  discoverer. 


GREGORY'S   SERIES.  239 

Sch.  These  results  may  be  applied  to  prove  any  general 
formula  in  elementary  Trigonometry,  and  are  of  'great  im- 
portance in  the  Higher  Mathematics. 

EXAMPLES. 

1.   Prove  Sm26>     =  tail  ft 

1  +  cos  20 


Prove  the  following,  by  the  exponential  values  of  the 
sine  and  cosine. 

2.  cos  2  a  =  cos2  a  —  sin2  a. 

3.  sin0  =  -sin(-0). 

4.  cos30  =  4cos30  —  3  cos  ft 

Rem.  —  If  we  omit  the  i  from  the  exponential  values  of  the  sine,  cosine,  and  tan- 
gent of  9,  the  results  are  called  respectively  the  hyperbolic  sine,  cosine,  and  tangent 
of  0,  and  are  written  sinh  0,  cosh  0,  and  tanh  9,  respectively.    Thus  we  have 
sinh  0  =  —  i  sin  iff,  cosh  0  =  cos  iff,  tanh  6  =  —  i  tan  iff. 

Hyperbolic  functions  are  so  called,  because  they  have  geometric  relations  with 
the  equilateral  hyperbola  analogous  to  those  between  the  circular  functions  and  the 
circle.  A  consideration  of  hyperbolic  functions  is  clearly  beyond  the  limits  of  this 
treatise. 

For  an  excellent  discussion  of  such  functions,  the  student  is  referred  to  such 
works  as  Casey's  Trigonometry,  Hobson's  Trigonometry,  Lock's  Higher  Trigonom- 
etry, etc. 


162.   Gregory's  Series.  —  To  expand  0  in  powers  of  tan  0 
lere  0  lies  between  —  -  and  +  -• 

By  (3)  of  Art.  161,  we  have 


1-f  t'tanfl 


9 \0i9 


240  PLANE  TRIGONOMETRY. 

.-.  log  e™=  log(l  -h  i  tan  0)  -  log(l  -  i  tan  (9). 


£tan50-etc.)    (Art.  130) 
n50-etc  .....     (1) 
which  is  Gregory's  Series. 

This  series  is  convergent  if  tan  6  =  or  <  1,  i.e.,  if  6  lies 
between  —  ^  and  -,  or  between  |TT  and  JTT. 


This  series  may  also  be  obtained  by  reverting  (5) 
in  Cor.  2,  Art.  156. 

Cor.  1.    If  tan  6  =  x,  we  have  from  (1) 

fam-ia?=*™4-~-etc  ......     (2) 

o       o 


Cor.  2.    If  0  =  -,  we  have  from  (1) 
4 


(3) 


a  series  which  is  very  slowly  convergent,  so  that  a  large 
number  of  terms  would  have  to  be  taken  to  calculate  TT  to  a 
close  approximation.  We  shall  therefore  show  how  series, 
which  are  more  rapidly  convergent,  may  be  obtained  from 
Gregory's  series. 

163.  Euler's  Series. 

tan-1  1  +  tan-1  1  =  £     .     .     .     (by  Ex.  2,  Art.  60) 

Put  0  =  tan-1  i.    A  tan  0  =  1,  which  in  (1)  of  Art.  162 
gives 


Put  e  =  tan-1  -•    .-.  tan  0  =  \,  and  (1)  becomes 
3      '  «3 

tau-1i=1--1-+-i  ___  L-4-etc  .....     (2) 
3      3     3.33^5-35      7-37 


MACHINES  SERIES.  241 

Adding  (1)  and  (2)  we  have 

=./i'lJL  J_    ..W3L_ii;  J_     A.-/** 

3  5  3  5 


4       2     3-23     5-25         /    \3     3.33     5.3 

a  series  which  converges  much  more  rapidly  than  (3)  of 
Art.  162. 

164.   Machin's  Series. 

Since  2  tan-1  -  =  tan-1  —  -i—  (by  Ex.  3,  Art.  60)=  tan"1  4> 
5  1—  12 


...^tan-'i-tan-'^. 
5~3~^P~5~.!?~ 


V239      3-(239)3  '  5  (239)5 
In  this  way  it  is  found  that  TT  =  3.141592653589793  •••. 


- 


- 


Cor.    Since  tan-      +  tan-        =  tan 


NOTE.  —  The  series  for  tan"1  —  and  tan"1  —  are  much  more  convenient  for  pur- 
70  99 

poses  of  numerical  calculation  than  the  series  for  tan"1  -- 

239 

Example.  —  Find  the  numerical  value  of  TT  to  6  figures  by 
Machin's  series. 


242  PLANE  TRIGONOMETRY. 

165.   Given  sin  0  =  x  sin  (0  +  «);  expand  0  in  a  Series  of 
Ascending  Powers  of  x. 

We  have    eie  —  e~i9  =  x[ei9+ia-  —  e~i9-ia-~\   .     .     (Art.  161) 

...    e2iO  -  1=  x     G2iO  .  eia  _  e-»a- 


.-.  2  10  =  log(l  -  xe~ia)  -  log(l  -  aeia) 

=  #(eia  -  6~ia)  4-  ^(e2ia_e-2ia)  _|_  ^ 

(Art.  130) 

.-.  ^  =  a;sma  +  -sin2«  +  -sin3«4-"-   (Art.  161)  (1) 
2  3 

Examine.   H  a  =  ir  —  20>  then  JB  =  1.     .-.  (1)  becomes 


166.  Given  tan  x  =  n  tan  ^  ;  expand  x  in  Powers  of  n. 

pix  _  p—  ix          aiO  _  p  -  id 

^  =  ne—  —  e—-    ....     (Art.  161) 

-ix  iO  __     -i9 


=  n-— 


/  1-|A 

[  where  m  =  —     - 
V 


1  4-  me2*0 
2io;  =  2^4-  log(l  +  me-**9)  —  log(l  4-  me2»fl) 


.  . 

.     .     (Art.  161) 


RESOLUTION  INTO  FACTORS.  243 

RESOLUTION  OF  EXPRESSIONS  INTO  FACTORS. 

167.  Resolve  xn  —  1  into  Factors. 
Since  cos  2  TIT  ±  V  —  1  sin  2  rir  =  1, 

where  r  is  any  integer,  and  xn  =  1, 

.-.  of  =  cos  2r7r  ±  V—  1  sin  2r7r. 


i 
.-.  a;  =  (cos  2?-7r  ±  V-  1  sin  2r»)" 


=  cos  —  ±  V^l  sin     E.     .     .     (Art.  153)  (1) 
n  n 

(1)   When  n  is  even.   If  r  =  0,  we  obtain  from  (1)  a  real 
root  1  ;  if  r  =  -,  we  obtain  a  real  root  —  1,  and  the  two  cor- 
responding factors  are  x  —  1  and  x  -f  1.     If  we  put 
,-  =  l,2,3...5-l, 

in  succession  in  (1),  we  obtain  n  —  2  additional  roots,  since 
each  value  of  r  gives  two  roots. 

The  product  of  the  two  factors,  which  are 


n  n  J 

and  '  -2- 

n  n 


n 

— 
n 

which  is  a  real  quadratic  factor. 


(2) 


-2a;  cos 

(3) 


*  This  expression  gives  the  n  nth  roots  of  unity. 


244  PLANE   TRIGONOMETRY. 

(2)     When  n  is  odd.     The  only  real  root  is  1,  found  by 
putting  r  =  0  in  (1)  ;  the  other  n  —  1  roots  are  found  by 

putting  r  =  1,  2,  3,  •••        —  in  (1)  or  (2)  in  succession. 


•  ••(  a;2—  2#cos-  —  Tr+l)(x2—2xcos-  —  rr+l..  (4) 


168.  Resolve  xn  +  l  into  Factors. 


Since     cos  (2r  +  !)*•  ±  V^T  sin  (2r  -f  I)TT  =  -  1, 
where  r  is  any  integer,  and  xn  =  —  1, 

.-.  af  =  cos(2r  +  l)7r±  V^l  sin  (2r  +  I)TT. 
.-.  a;  =[cos(2r-f  l)ir  ±V:rI  sin  (2r  +  !)*•]" 


/  —  T    .    2r+l 

=  cos  -  7T±V—  Ism—     —  TT  .     .     .     (1) 
n  n 

which  is  a  root  of  the  equation  xn  =  —  1  ;  ie.,  —  1  is  a  root. 

(1)    When  n  is  even.     There  is  no  real  root  ;  the  n  roots 
are  all  imaginary,  and  are  found  by  putting 


successively,  in  (1). 

The  product  of  the  two  factors, 


and 

=  x*-2xcos^-±*7r  +  l  .     (2) 

n 

which  is  a  real  quadratic  factor. 


RESOLUTION  INTO  FACTORS.  245 


STT 
n 


(3) 


(2)    When  n  is  odd.     The   only  real   root   is    —  1 ;   the 

other  n  —  1  roots  are  found  by  putting  r  =  0,  1,  2,  •  •  •  — 
in  (1),  in  succession. 

.-.  xn  + 1=  (x  +  1/ar2  -  2acos  -  +  lYo2  -  2»  cos  —  +  1 


" 


(4) 


EXAMPLES. 


1.  Find  the  roots  of  the  equation  x5  —  1  =  0. 

Ans.  1,  cos  £(2nr)-M'sin|(2rir),  where  r=l,  2,  3,  4. 

2.  Find  the  quadratic  factors  of  as8  —  1. 

.4ns.  (a2  -  1  )  (y?  -  V2  a;  +  1)  (a?  +  1)  (aj2  +  V2  a;  +  1)  . 

3.  Find  the  roots  of  the  equation  »4  +  1  =  0,  and  write 
down  the  quadratic  factors  of  x*  +  1- 

Ana.  ±  -L  ±  V^l  A:;  (a?  -  W2  +  1)  (x-2  +  W2  +  1). 
V2  V2 

169.  Resolve  iB271  —  2  #M  cos  ^  +  1  into  Factors. 

Let   a^M-2a;ncos^  +  l=:0. 
...  z?n  —  2xn  cos  0  +  cos2  0  =  —  sin2  6. 


.-.  xn  —  cos  0  =  ±  V—  1  sin  0  =  ±  i  sin  0. 

/.  a;  =  (cos  B±  i  sin  0)"  =  cos  2r7r  +  °  ±  i  sin  2r7r  +  B    (1) 

n  n 

since  cos  0  is  unaltered  if  for  0  we  put  0  +  2  r?r.  If  we  put 
r  =  0,  1,  2,  -"n  —  1,  successively  in  (1),  we  find  2w  differ- 
ent roots,  since  each  value  of  r  gives  two  roots. 


246  PLANE  TKlGONOMETItY. 

The  product  of  the  two  factors  in  (1) 

/  2nr  +  0      .   . 

=   x  —  cos ~ i  sin 


n       J 

sin    ?  v  "*" 


n 

•i GO 

n 

.-.  x2n  —  2xncosO  +  l 


=  for2  -  2  a  cos  -  + 1  Yor2  -  2  z  cos  2?r  +  ^  +  A . . 
V  w        A  w  ) 

/                     (271-4)^  +  0        \ 
...f  a;2  —  2  a;  cos ^ hi  )••• 

.. (3) 


Cor.   Change  x  into  -  in  (3)  and  clear  of  fractions,  and 
we  get        x2n  -  2  anxn  cos  0  +  a2"  =  (a?  -  2  ax  cos-  +  a2V- 

•  •  •(  x2  —  2  ax  cos    v     — [-  a2 }( x2  —  2  ax  cos    ^    — |-  a2 
^  n  J\  n 

•  ••to  n  factors (4) 

EXAMPLES. 

Find  the  quadratic  factors  of  the  following : 

Ans.  (x*-2x  cos  15°  +  1) (x2  -  2  x  cos  105°  +  1) 

X  (x2  -  2  x  cos  195°  +  1)  O2  -  2  x  cos  285°  -f  1)  =  0. 

2.   x10- 2^00810°  +  1=0. 

Ans.  (x*-2x cos 2°  +  1)  (or2 -  2x cos 74°  +  1) 
X  (x2  -  2  x  cos  146°  +  1)  (x2  -  2  a;  cos  218°  -f  1) 
(ar'  -  2  x  cos  290°  -f  1)  =  0. 


DE  MOIVRWS  PROPERTY  OF  THE  CIRCLE.      247 


170.  De  Moivre's  Property  of  the  Circle.  —  Let  0  be  the 

centre  of  a  circle,  P  any  point  in  its 

plane.     Divide  the  circumference  into  D 

n  equal  parts  EC,  CD,  DE,  •••,  begin- 

ning  at  any  point  B;  and  join  0  and 

P  with  the   points  of   division  B,  G, 

D,  ....     Let  POB  =  6;  then  will 

OB2M-2  OB"  •  OP"  cos  ?i0+OP2n 
=  PB2  •  PC2  •  PD2.  •  •  to  n  terms. 

For,  put  OB  =  a,  OP  =  a;,  and  0  =  -;  then 


PB2  =  OP2  +  OB2  -  2  OP  .  OB  cos  0 


=  x-  +  a2  —  2  ax  cos  - 
n 


(1) 


PC2  =  OP2  +  OO2  -  2 OP  •  OC  cos 


=  x2  -f  a2  —  2  ax  cos  —  — — — ;  and  so  on 

n 

Multiplying  (1),  (2),  (3),«"  together,  we  have 


PB2  •  PC2  •  PD2  •  •  •  to  n  terms 


=  [or  —  2  ax  cos  -  + 
n 


—    ax  cos 


-  2  ax  cos 


+  a2 


=  a2n  -  2  a"^  cos  «  +  a2"    .     [by  (4)  of  Art.  169] 

B2n     ...     (3) 


which  proves  the  proposition. 


248  PLANE  TRIGONOMETRY. 

171.   Cote's  Properties  of  the  Circle.  —  These  are  particu- 
lar cases  of  De  Moivre's  property  of  c     C 
the  circle. 

(1)   Let  OP,  produced  if  necessary, 
meet  the  circle  at  A,  and  let 


n 

then  nO  is  a  multiple  of  2?r.     Hence 
we  have  from  (3)  of  Art.  170,  after  taking  the  square  root 
of  both  members, 

OB"  _  OF  =  PB  .  PC  •  PD  ...  to  n  factors  ...     I. 

(2)  Let  the  arcs  AB,  BC,  ---be  bisected  in  the   points 
a,  &,•••;  then  we  have,  by  (1), 

OB2M  _  OP2'1  =  Pa  •  PB  •  P6  .  PC  •  PC  •  .  •  to  2  n  factors. 
Hence,  by  division, 

OBn  +  OPn  =  Pa  •  P6  •  PC  ...  to  n  factors  ...     II. 

Cor.    If  the  arcs  AB,  BC,  ...  be  trisected  in  the  points 
(*!,  a2,  61,  62,  *  *  •>  then  we  have 
OB2n  +  OBM.OPn+OP2n=Pa1.Pa2.P61.P62...  to  2n  factors. 

172.  Resolve  sin  &  into  Factors. 

(1)  Put  x  =  1  ;  then  we  get  from  (3)  of  Art.  169 


(1) 


\ 

Put  0  =  2n<f>  in  (1),  and  let  2na  = 

1  —  cos  0  =  1  —  cos  2n<f>  =  2  sin2w</> ; 
then  extracting  the  square  root,  we  have 

i  =  2n~1sin<£  «  sin  (<£  +  2«)  sin  (<£  +  4 a)  x 

•  •  x  sin  (<£  +  2na  —  2  a) (2) 


RESOLUTION  INTO  FACTORS.  249 

But  sin  (<f>  +  2  na  —  2  a)  =  sin  (<£  +  TT  —  2  a)  =  sin  (2  «  —  <£)  , 
sin  (<£  -f-  2  n«  —  4  «)  =  sin  (4  «  —  <£),  and  so  on. 

Hence,  when  n  is  odd,  multiplying  together  the  second 
factor  and  the  last,  the  third  and  the  last  but  one,  and  so 
on,  we  have 

sin  n<j> 

=  2n~l  sin  </>  sin  (2  a  +  <£)  sin  (2  a  —  <f>)  sin  (4  a  -f  <f>)  sin  (4  a  —  <f>) 
•  ••  x  sin  [(w  —  1)«  +  <£]  sin  [(??  —  1)«  —  <^]. 

But  sin  (2  «  +  </>)  sin  (2  a  —  <j>)  =  sin3  2  a  —  sin2  <£,  and  so  on. 

.-.  sinn<£=2n~1sin<£(sin22«  —  sin2(/>)  (sin24«  —  sin2<^>)  x  ••• 
...  X  [sin2(?i  —  1)«  —  sin2<^>]     ....     (3) 

Divide  both  members  of  (3)  by  sin  <£,  and  then  diminish 
<£  indefinitely.  Since  the  limit  of  sin  n<fr  -?-  sin  <f>  is  n,  we 
get 

w  =  2w-1sin22asin24asin26«x  •-«  xsin2(w  —  1)«     (4) 

Divide  (3)  by  (4)  ;  thus 

x...      (5) 


Put  7i^>  =  0,  and  let  n  be  increased  while  <£  is  diminished 
without  limit,  0  remaining  unchanged  ;  then  since  2  na  =  TT, 
the  limit  of 


=  -*  (Art. 


sin  -  -          sm  - 

n  \n  n 


/i 

and  the  limit  of  n  sin  <£  =  that  of  n  sin  -  =  0;  and  so  on. 

n 

Hence  (5)  becomes 


NOTE.  —  The  same  result  will  be  obtained  if  we  suppose  n  even. 


250  PLANE  TRIGONOMETRY. 

Hem.  —  When  0>0  and  <  TT,  sin  9  is  +,  and  every  factor  in  the  second  member 
of  (6)  is  positive;  when  0>n-  and  <2n,  sin  6  is  — ,  and  only  the  second  factor  is 
negative;  when  0>2  TT  and  <3n-,  both  members  are  positive,  since  only  the  second 
and  third  factors  are  negative ;  and  so  on.  Hence  the  +  sign  was  taken  in  extract- 
ing the  square  root  of  (1). 

Cor.  Let  0=-,  then  sin  -  =  1,  and  -  =  1.  Hence  (6)  be- 
comes 


7T 


2     22 
22       42 


2      1-3  3-5  5-7  7-9 
which  is  Wall-is? s  expression  for  TT. 

173.   Resolve  cos<9  into  Factors.  — In   (2)  of  Art.  172, 

change  <£  into  <f>  -\-  a,  then  n<j>  becomes  n<j>  +  nit,  i.e.,  w<£  +  -• 
Hence  (2)  becomes 

But  sin(<£  -f  2n«  —  a)  =  sin(<£  -f-  TT  —  a)  =  sin  (a  —  <£), 

sin(^)  +  2w«  —  3«)==  sin(3«  —  <^>),  and  so  on. 
Hence  when  n  is  even  we  have  from  (1) 
cos  n<£=2"~1  sin(«  +  ^>)  sin(«  —  <^>)  sin(3«  +  ^>)  sin(3«  —  <£) 

=  2n-](sin2  a  —  sin2<£)  (sin23«  —  sin2^) 

X  •••  x[sin2(w-l)«-sin2<£] (2) 

Therefore,  putting  n<f>  =  0,  as  in  Art.  172,  we  obtain 

-$£\ (3) 


NOTB.  — For  an  alternative  proof  of  the  propositions  of  Arts.  172  and  173,  see 
Lock's  Higher  Trigonometry,  pp.  92-95. 


SUMMATION  OF  SERIES.  251 

EXAMPLES. 

1.  If  «  =  — ,  prove  that 

4  n 

sin  a  sin  5  a  sin  9  a  •  •  •  sin  (4  n  —  3)  a  =  2~n+i 

2.  Show  that 


=  cos  5  9. 
SUMMATION    OF    TRIGONOMETRIC    SERIES. 

174.  Sum  the  Series 
We  have 


2  sin  a  sin  J/?  =  cos    a  —  ^ )—  cos  (  a  -\-  ^  )  (Art.  45) 
y     r2/          \       v 

2  sin  («  +  /?)  sin i)8  =  cos  (a  +  %)-  cos  («  +  f  0), 
\        V 

2  sin  (a  +  2^8)  siniyg  =  cos  («  +  f  ft)  -  cos  («  + 

etc.  =  etc. 
2  sin  [«  +  (w  —  I)/?]  sin  $fi 

=  cos|-«  +  2^V]  -  eos[«  +  ?^l 

Therefore,  if  Sn  denote  the  sum  of  n  terms,  we  have,  by 
addition, 

2Sn  sin  %f3  =  cos  («  -  £0)  -  cosf«  +  27l~1 

+  5— ^/glsiniwjg  .     .    (Art.  45) 


252  PLANE  TRIGONOMETRY. 

175.   Sum  the  Series 

cos  «  +  cos  («  +  /?)  +  cos  («  +  2/?)H  -----  hcos[a+(-n  —  I)/?]. 
We  have  2  cos  a  sin  J/3  =  sin  (a  +  ^-/J)  —  sin  (a  —  %(3), 
2  cos  («  +  /?)  sin  i0  =  sin  («  +  f  /?)  -  sin  («  +  £/3), 

etc.  =  etc. 
2  cos  [«  +  (71  —  1)£]  sin  £  /? 


Denoting  the  sum  of  n  terms  by  Sn,  and  adding,  we  get 
2  SB  sin  £0  =  sin  L  +  ^=11  -sin    «- 


cos  fa  +  ^^^  pl  sin  |  w 

.         -_L__±_J 


—  The  sum  of  the  series  in  this  article  may  be  deduced  from  that  in  Art.  174 
by  putting  a  +  -  for  a.  The  sums  of  these  two  series  are  often  useful;*  and  the 
student  is  advised  to  commit  them  to  memory. 

Cor.    If  we  put  /?  =  —  ,  then  sin  $nfi  —  sin  TT  =  0.    Hence 
we  have  from  Arts.  174  and  175 


n 

NOTE.  —  These  two  results  are  very  important,  and  the  student  should  carefully 
notice  them. 

176.   Sum  the  Series 


This  may  be  done  by  the  aid  of  Art.  159  or  Art.  160. 

*  See  Thompson's  Dynamo-Electric  Machinery.    3d  ed.,  pp.  345,  346. 


SUMMATION  OF  SERIES.  253 

Thus,  if  m  is  even,  we  have  from  Art.  159 
2m~1  sinm  «  =  (  —  l)2[cosm«  —  m  cos  (m  —  2)«H  ----  ]     .     (1) 

2m-l  ginm  (a  +  £) 
m 

=  (-If  [cos  m  (a+p)  -m  cos  (m-2)  («+£)  +  ...]       (2) 

and  so  on  ;  and  the  required  sum  may  be  obtained  from  the 
known  sum  of  the  series 

[cos  ma  +  cos  m  (a  +  (3)  4-cosra  («-f2/3)  H  ----  ] 
and       Jcos(ra  —  2)  <*  +  cos[(m  —  2)  (<*  +  /?)] 

+  cos  [(m  _  2)  (a  +  2  j8)]  +  ,,-j,  etc. 
We  may  find  the  sum  of  the  series 

cosm  «  +  cosm  («  +  p)  +  cosm  («  +  2  /?)  +  etc. 
to  n  terms  in  a  similar  manner  by  the  aid  of  Art.  158. 

EXAMPLES. 

1.    Sum  to  n  terms  the  series 

sin2  a  +  sin2  («  +  £)  +  sin2  (a  +  2  /3)  +  ••• 

We  have 

2  sin2  a  =  -  (cos  2  a  -  1)  by  (1), 

2  sin2(«  +  P)  =  -  [cos  2  (a  +  /?)  -  1]  by  (2), 

2  sin2  (a  +  2  £)  =  -  [cos  2  (a  +  2  /3)  —  1],  and  so  on. 

Hence 

2Sn=tt-[cos2a+cos2(«+/?)+cos2(«+2/?)  +  ..-] 
=  n  -  cos[2«  +  (n 


=n      cos  [2  «  +  (n  —  1)^8]  sin  yi^ 
~2 


254 


PLANE  TRIGONOMETRY. 


2.    Sum  to  n  terms  the  series 

cos3  a  +  cos3  2  a  +  cos3  3  a  +  •  •  • 

2  cos  [3  a  +  k  (n  —  1)  3  «]  sin 


8  sin  £  a 


6  cos   a 


-j 


sin 


wet 


8  sin  i  a 
177.   Sum  the  Series 

sin  a  —  sin  («  +  /?)  +  sin  («-f-  2  /?) to  n  terms    .      (1) 

Change  /?  into  (3  +  TT,  and  (1)  becomes 

sina  +  sin(«  +  7T  + j8)  +  sin(«  +  27r-l-2^)H .     .     (2) 

Therefore  we  have  from  Art.  174 

Sill 


inf.. +  ("!.!)( 


Similarly, 
cos«  —  co 


----  to  n  terms 


8n 


178.   Sum  the  Series 

cosec  6  +  cosec  2  0  -f  cosec  4  0  -f-  •  •  •  to  n  terms. 

f\ 
We  have         cosec  0  =  cot  --  cot  0, 

2 

cosec  2  0  =  cot  0  —  cot  26, 

etc.  =  etc. 

cosec  2"-10  =  cot  2n~20  -  cot  22n~10. 
Therefore,  by  addition,  as  in  Art.  174, 


SUMMATION  OF  SERIES.  255 

NOTE.  —  The  artifice  employed  in  this  Art.,  of  resolving  each  term  into  the  dif- 
ference of  two  others,  is  extensively  used  in  the  summation  of  series. 

Practice  alone  will  give  the  student  readiness  in  effecting  such  transformations. 
If  he  cannot  discover  the  mode  of  resolution  in  any  example,  he  will  often  easily 
recognize  it  when  he  sees  the  result  of  summation. 

The  student,  however,  is  advised  to  resort  to  this  method  of  solution  only  as  a  last 
resource. 

179.   Sum  the  Series 

n  /a 

tan  0  H-  ^  tanr  +  \  tan-  -f  •••  to  n  terms. 
2  4 

We  have          tan0  =  cot0-  2  cot  20, 
|tan-  =  icot^ 


etc  =  etc. 

_  tan  —  =  —  cot  —  ___  —  cot 

»-i          "-1       n~l         "-1       n~2         n~ 


•-  i 

180.  Sum  the  Series 


Denote  the  sum  by  SM,  and  substitute  for  the  sines  their 
exponential  values  (Art.  161).     Thus, 


xn-l  j-gt(a  +  n/3-/3)  _g-t(a  +  n/3-^)  J 


_^_  xn+l  |^et(a+n)3-j3)  _  g-i(a+n/3-p)  J 


256  PLANE  TRIGONOMETRY. 


sing—  a?sin(«—  ft)—  ansin(ft+n/3)+af+1sin[tt+(?i  —  I)/?] 


Cor.   If  a;  <  1,  and  ?i  be  indefinitely  increased, 

g    _  sin  a  —  x  sin  (a  —  ft) 
1  —  2XCOSP  +  X2 

Sch.    Similarly, 

cos  a  -f  x  cos  («  +  /3)  +  x2  cos  (a  +  2  ft)  -f-  •  •  •  to  w  terms  = 
cos  a  —  x  eos(  ct  —ft)  —  xn  cos(  a  +  nft)  +  &n+1  cos  [«  +  (^  — 


/  o  \ 

We  may  obtain  (3)  from  (1)  by  changing  a  to  a  -f-  ^« 

2 

Also  Soo  =  cosa-a;cos(tt-/3)  (4) 

1  —  2xcos^  +  a;2 

181.  Sum  the  Infinite  Series 

a;  sin  («  +  £)  +  ^sin  («  +  2/3)  +  ^sin  («  +  3j8)  +  ••., 

I?  Le 


and 


Let  S  denote  the  former  series,  and  C  the  latter. 


Then  C  +  £S  =  xe 


6*«(e*«*  -  1)      .     .    .[by  (3)  of  Art.  129] 


_l-)       .      .      .       (Art  161) 

gX  cos  /3g  i(a  +  iC  sin  /3)  _  gta 


—  (cos  a  -f-  *  sin  a)     ....     (Art.  161) 


EXAMPLES.  257 

Equating  real  and  imaginary  parts,  we  have 
C  =  exco*P  cos  (a  -f  x  sin  /?)  —  cos  a, 
S  =  excosP  sin  (a  -f-  x  sin  /3)  —  sin  a. 

EXAMPLES. 
Prove  the  following  statements  : 

1.  The  two  values  of  (cos  4  0  +  V^T  sin  4  0)*  are 

±  (cos  20  +  V^I  sin  2  0)      ...     (Art.  154) 

2.  The  three  values  of  (cos  0  +  V  —  1  sin  0)  i  are 

6     ,         /  -  T      •       0  27T  +  0     ,         /  -  ?      •       27T  +  0 

COS-+V  —  1  smr,   cos  —  r:!:  —  hv—  Ism  -  ^—  , 
3  o  o  3 

47T  +  ^     .         /  -  ?      •       47T  +  0 

cos  -  :!:  —  h  V  —  1  sin  -  JE  — 

3.  The  three  values  of  (—1)^  are 

l+f*.    -1,    ±f*     .....     (Art.  154) 

4.  The  six  values  of    —1^  are  contained  in 


cos  ±  Vl  sin  ,  where  r  =  0,  1,  or  2. 

6  6 

5.   The  three  values  of  (1  -f  V—  I)*  are  contained  in 
^:lsinfl  where  0  =  j,  JTT,  or-V^. 


6 

6.    The  three  values  of  (3  +  4V—  !)•*  are  contained  in 
+  V^l  sin         ±^1  where  r  =  0,  1,  or  2. 


7.  cos  6  0  =  cos6  0-15  cos4  0  sin2  0  +  15  cos2  0  sin4  0  -  sin6  0. 

8.  sin  90=9  cos8  0  sin  0-84  cos6  0  sin3  0+126  cos4  0  sin5  0 

-36cos20sin70  +  sin90. 


258  PLANE  TRIGONOMETRY 

9.   tan  nO 


[2 

10.  Given  -  =  -  :  show  that  0  is  nearly  the  circular 

B        2166 
measure  of  3°. 

Prove  the  following  : 

11.  -  64  sin7  0  =  sin  7  0  —  1  sin  5  6  +  21  sin  3  0  —  35  sin  0. 

12.  -29sin100=cos  10  0-10cos  8  0  +  45  cos  6(9-120  cos  40 

+  210  cos  2  0  -  126. 

13.  26  (cos8  0  +  sin8  0)  =  cos  8  0  +  28  cos  4  <9  -f  35. 

14.  cos6  <9  +  smfi0  =  -|(5  +  3  cos  40). 

15.  Expand  (sin0)4'l+2  in  terms  of  cosines  of  multiples 
of  0. 

16.  Expand  (sin  0)4n+1  in  terms  of  sines  of  multiples  of  0. 

17.  Expand  (cos  0)2n  in  terms  of  cosines  of  multiples  of  0. 

Use  the  exponential  values  of   the  sine  and  cosine  to 
prove  the  following  : 


18. 


1  -  cos  0  2 

19.    If  log  (x  4-  y  V^T)  =  a  +  (3  V—  1,  prove  that 
cc2  -f  i/2  =  e2a,    and   y  =  x  tan  /?. 


20.   If  sin(«  +  £V—  l)  =  a;  +  y  V—  1,  prove  that 
x2  cosec2  a  —  y2  sec2  «  =  1. 

21. 


EXAMPLES.  259 

7T 

22.      V^i-^e^2. 


23.  e'(cos  (9  +  V^I  sin  0)  =  eV    cos      .  V^T  sin  - 

\       4  4/ 

24.  The  coefficients  of  xn  in  the  expansion,  (1)  of  e°*  cos  &#, 
and  (2)  of  eax  sin  &#,  in  powers  of  #,  are 


25.  The  coefficient  of  af*  in  the  expansion  of  e*  cos  a;  in 

n 
02  ^ 

powers  of  x  is  —  cos  —  -• 

[n          4 

26.  If  the  sides  of  a  right  triangle  are  49  and  51,  then 
the   angles   opposite   them   are   43°  51'  15"  and   46°  8'  45" 
nearly. 

27.  If  a  and  b  be  the  sides  of  a  triangle,  A  and  B  the 
opposite  angles,  then  will  log  6  —  log  a 

=cos  2  A  —  cos  2B  +  ^(cos  4  A  —  cos  4B) 

—  cos  6B)  H  ----  . 


28.  If  A  -f  iB  =  log(m  +  m),  then 

tan  B  =  -,  and  2  A  =  log  (n2  +  m2) . 

29.  cos  (0  4-  to)  =  cos  0 

30.  sin  (0  + i\M  = 


V*  >  /  i  *  > 

J       /  \       J 

31.    2  cos  (a  -f- «'/?)  =  cos  «(e-9  +  e~^)  —  i  sin  «(e^  —  e-0). 

—  rae  ~^r[cos  (^8  log  r  +  a?*)  -f-  i  sin  (/?  log  r  -f  ar)  ], 
where   a  +  ^  =  ?*(cos  r  -\-i  sin  r) . 

33.    log  (a  +  ib)  =  4  log  (a2  -f  62)  4-  i  tan-1  ^ . 

a 


260  PLANE  TRIGONOMETRY. 

34. 


=  sin'^afsinfa  —  nO)  +  eiia  sin  n0]. 


35.  |  =  _i-  +  -i-  +  _L-  +  -'-. 

36.  Write  down  the  quadratic  factors  of  a;13  —  1. 

Ans.    (x  —  1)  [cc2  —  2  #  cos  Jg  (2  r?r)  -f  1],  six  factors, 
putting  r  =  1,  2,  3,  4,  5,  6. 

37.  Solve  the  equation  x&  —  1  =  0. 


38.  Give  the  general  quadratic  factor  of  x20  —  a20. 

-4?is.  #2  —  2  ax  cos  y1^  (rw)  +  a2. 

39.  Find  all  the  values  of  A/1. 

^4ns.  cos  i(r7r)4- 1  sini(?-7r),    r  having    each   integral 
value  from  0  to  11. 

40.  Write  down  the  quadratic  factors  of  #6  -j-  1. 

Ana.     x2-V3o; 


41.  Write  down  the  general  quadratic  factor  of  x®  -f  1. 

Ans.  x2  —  2  x  cos  (1  +  2  r)  9°  +  1. 

42.  Find  the  factors  of  a13  +  1  =  0. 

Ans.   (x  +  1)  [#2  —  2  x  cos  T^  (TT  +  2  PTT)  +  1],  seven  fac- 
tors in  all. 

43.  Find  a  general  expression  for  all  the  values  of  -\/—  1. 

W  -h  2  T'TT  •       7T  -4-  2  ?*7T 

^bis.  cos — |-  i  sin  —  -I— - — ,    where   r  may   have 

n  n 

any  integral  value. 

44.  Solve  x12- 2 0^0081^  + 1  =  0. 

Ans.  x2  —  2  x  cos  j-  (3  r-n-  +  TT)  +  1  =  0,  six  quadratics. 


EXAMPLES.  261 


45.  Solve  z10  +  V3  x>  +  1  =  0. 

Ans.  x2  +  2x  cos  (r  x  72°  +  6°)  +  1  =  0,  five  quadratics. 

46.  Write  down  the  quadratic  factors  of 

yan  _  2  xnyn  cos  a  +  y~n. 


Ans.  y?-2xy  cos  "  +  y\  u  factors. 


Prove  the  following  : 

47.   tan^tan^  +  ^tan-h—    •  •  tan       + 


(—1)2,  where  n  is  even.     [Use  (2)  of  Art.  172.] 

48.    sin  5  0  -  cos  5  0  =  16  cos  (0  -  27°)  cos  (0  +  9°)  x 

X  sin  (6  +  27°)  sin  (0  -  9°)  (cos  0  -  sin  0). 


50.   --.-. 


^1       —  -4-        _|_    L  _L  —  4-  . . .  —  — . 

'    12  +  22  +  32  +  42  ~6 

52.   1  +  1+1  +  1  +  ...  =  !?. 

I2     32      52     72  8 

53        =3    36    144   324   576 

'   *" "    ' '  35  '  143  '  323  '  575  ' 


54    ir  = 


2      1.3.3.5.5.7.7.9- 


55       /o^. 
" 


3.35.    99.195-323 
8.  80-  224.  440... 


262  PLANE  TRIGONOMETRY. 

57.   cos  x  +  tan    sin  x  — 


58.   cos  x  —  cot    sin  x  = 


59.  By  aid  of  the  formula  cos  0  =  S1]         and  Art.  172, 

2sin0 

deduce  the  value  for  cos  6  obtained  in  Art.  173. 

60.  By  expanding  both  sides  of  Ex.  57  in  powers  of  x 
and  equating  the  coefficients  of  x,  prove  that 


L     TT — y     7T+2/     oir — y     o7r-\-y     OTT — y 
61.   Prove  in  like  manner  from  Ex.  58  that 

29999 
&  £i  .  .     £  •    6 


2       y       2-rr-y       2>rr 


63.   Prove  =  1--      +      -+ 


64.   Prove  that  -J- = 


_ 

y     7r-y    2-rr-y    tr+y    2-r+y    STT—  y    ±ir-y    3-r+y 
Sum  the  following  series  to  n  terms  : 


65.    sinaH-sin2«+sin3aH  -----  \-s'mna  = 


sn 


EXAMPLES.  263 


66.  cos  a  +  cos  2  a  +  cos  3  a  -\  -----  h  cos  na  = 

nrr 

67.  si 

68. 


sn 

sin2?i« 

—     —  • 

sin« 

sm      "• 

2  sin  a 


nn      .o        .   90         .   20  wsin«  —  sinn«cos(?i 

69.    sm2tt  +  sm22«+sm23«H  —  =— 

2sm« 


70. 


2 -sin  a 
71.    sin3«  -f  sin3(«  +  ft)  +  sin3(«  +  2/3)  +  ... 


72. 


73.    sin  a  sin  2  a  -f-  sin  2  a  sin  3  «  +  sin  3  a  sin  4  «  4-  •  •  • 

_  n  sin  a  cos  «  —  sin  na  cos  (n  +  2)« 
2  sin  a 


74.  tan«-f-2tan2tt  +  22tan22«H  ----  =  cot  «  —  2n  cot  2n  «. 

75.  (tan  «  +  cot  a)  +  (tan  2  a  +  cot  2  «)  +  (tan  22«  -f  cot  22«) 

H  ----  =  2  cot  a  —  2  cot  2W  a. 

76.  sec  a  sec  2  a  4-  sec  2  a  sec  3  a  +  ••• 

=  cosec  a  [tan  (n  +  1)  a  —  tan  «]  . 


264  PLANE  TRIGONOMETRY. 

77.    cosec  a  cosec  2  a  +  cosec  2  «  cosec  3  a  -\  ---- 

=  cosec  a  [cot  a  —  cot  (?i  +  1)  «]. 

78         sin20  sin40        ...  ,.._.  sec  (2  71  +  1)0-  sec  0 

cos  0  cos  3  6      cos  3  0  cos  5  0  2  sin  0 

79.    cos4  a  +  cos4  («  +  /?)  +  cos4  (<*  +  2/3)  +  ...  =  f  n  + 


cos[2«  +  (n  —  l)/3]sinyiff     cos  [4«  +  (K  —  1)2/8]  sin  2n 
2  sin  0  8  sin  2  /? 

80    t  n  nO  _  sin  ^  +  sin  3  0  +  sin  5  0  H  ----  to  n  terms 
cos  0  +  cos  30  +  cos  5  0  -f-  •  •  •  to  n  terms 

81.    cos  $  cos(0  +  a)  +  cos(0  +  «)cos(0  -f  2  a) 

+  cos(0  +  2  a)cos((9  +  3  a)  +  • 

=  ^  cos  a  + 


2  2  sin  « 

sin  0—  sin  20+sin  30  ----  to  91  terms 


cos  0—  cos20+cos30  ----  to  n  terms 


=tan 


83.  sm(p  +  1)0  cos  0  +  sin(p  +  2)0  cos  2  0  +  ... 

sin  ( p  +  1  +  ?i )  0  sin  ?i0 
2sin0 

84.  sin  3  0  sin  0  +  sin  6 0  sin  2  0  +  sin  12  0  sin  4  0  +  ... 

=  1  (cos  20 -cos  2^0), 

85.  sin  0  (sin- ) +2  sin- (sin  — ) +4  sin    (sin 

=  2n~2  sin  -~  -  J  sin  2  0. 

A  A          A  A          A  A 

86.  tan    sec0  +  tan- sec -  +  tan-sec-  +  ."=tan  0—  tan—. 

2  4284  2n 

87.  cot  0  cosec  0+2  cot  2 0  cosec  2  0+22  cot  220  cosec  220+  ••• 


EXAMPLES.  265 


oo  -L  i -*• I -*- j 

0  sin  20      siu~20  sin  3  0      sin  3  0  sin  4  0 


sin 


=  --  -(cot  30 -cot  40). 
sm0v 


89. 


sin  0  cos  2  0      cos  2  0  sin  30      sin  3  0  cos  4  0 

=  cose\   •  2,L     . 
1 


(0  +  |Y  tan  (n  +  1)  f$  +  |V  tai/0  + |YI 


90.    tan-1 


91. 


1    _l_  ^  _l_  ^2 

l  -t-o  4-  o 

-E«t  n-1— i- 
~4        aU    n  + 


=  tan"1  nx. 


92.    sin  a  sin  3  a,  +  sin  -  sin  — -  +  sin  —^  sin  — " 


93.  -H  --  -  --  1  --  _  + 

cos  0  +  cos  30      cos  0  +  cos  5  0      cos  0  +  cos  7  0 


=  £  cosec  0  [tan  (  w  +  1  )  0  —  tan  0]  . 

>sec20sec220+... 

=  sin0(cot0  —  cot2"0). 


94.    -sec0  +  isec0sec20-hisec0sec20sec220+  •• 

22  2 


95. 


=  log  2  sin  2  0  -      log  2  sin  2n+1  0. 
Sum  the  following  series  to  infinity  : 


96. 


[2  [3 

=  eco*2e  cos  (0  +  sin  0  cos  0). 


266  PLANE  TRIGONOMETRY. 


If 

98.     l-C 


99.   2  cos  0  +  1  cos20  4-  1  cos30  +  £  cos4l9  -f 


cos  6 

=  -        —  -Iog(l-cos0). 
1  —  cos  0 


100.    s 


[2  [S 

"T"    '  [2  ' 

L-  i  i  i    /Q  n 

102.   si 


|2 

==  6  sin  ^p  4~  sin  (/)• 

103.  cos0-|cos204-icos30 =  log (% cos -Y 

104.  cos  20  +  ^  cos  60  +  I  cos  100  4  ...  =  ilog  cot-. 


105. 


106.  x  cos  0  -  -  cos  2  0  4-  ~  cos  3  0  -  -  cos  4  0  +  . . . 

234 

=  log(l  4-  2x  cos0  4-  x'J. 

107.  sin  0  sin^  _  sin  2  0 ^5_?  4.  sin 3 0  -^ 

=  cot"^!  4-  cot20  4-  cot0). 

108.  -  +  i  +  1  +  1+...=^!. 
I4      24      34      44  90 

109.  1  a.  i  _|-  1  4. 1  _i =?*-. 

I4     34     54      74  96 


PART  II. 
SPHERICAL    TRIGONOMETRY. 

CHAPTER   X. 

FORMULA  EELATIVE  TO  SPHEKIOAL  TRIANGLES, 

182.  Spherical  Trigonometry  has  for  its  object  the  solu- 
tion of  spherical  triangles. 

A  spherical  triangle  is  the  figure  formed  by  joining  any 
three  points  on  the  surface  of  a  sphere  by  arcs  of  great 
circles.  The  three  points  are  called  the  vertices  of  the 
triangle ;  the  three  arcs  are  called  the  sides  of  the  triangle. 

Any  two  points  on  the  surface  of  a  sphere  can  be  joined 
by  two  distinct  arcs,  which  together  make  up  a  great 
circle  passing  through  the  points.  Hence,  when  the  points 
are  not  diametrically  opposite,  these  arcs  are  unequal,  one 
of  them  being  less,  the  other  greater,  than  180?.  It  is  not 
necessary  to  consider  triangles  in  which  a  side  is  greater 
than  180°,  since  we  may  always  replace  such  a  side  by  the 
remaining  arc  of  the  great  circle  to  which  it  belongs. 

183.  Geometric   Principles.  —  It   is   shown  in  geometry 
(Art.  702),  that  if  the  vertex  of  a  triedral  angle  is  made 
the  centre  of  a  sphere,  then  the  planes  which  form  the 
triedral  angle  will  cut  the  surface  of  the  sphere  in  three 
arcs  of  great  circles,  forming  a  spherical  triangle. 

Thus,  let  0  be  the  vertex  of  a  triedral  angle,  and  AOB, 
BOG,  COA  its  face-angles.  We  may  construct  a  sphere 
with  its  centre  at  0,  and  with  any  radius  OA.  Let  AB, 

267 


268  SPHERICAL    TRIGONOMETRY. 

BC,  CA  be  the  arcs  of  great  circles  in  which  the  planes  of 

the  face-angles  AOB,  BOG,  COA 

cut  the  surface  of  this  sphere ; 

then  ABC  is  a  spherical  triangle, 

and  the  arcs  AB,  BC,  CA  are  its 

sides. 

Now  it  is  shown  in  geometry 
that  the  three  face-angles  AOB, 
BOC,  COA  are  measured  by  the  sides  AB,  BC,  CA,  re- 
spectively, of  the  spherical  triangle,  and  that  the  diedral 
angles  OA,  OB,  OC  are  equal  to  the  angles  A,  B,  C,  respect- 
ively, of  the  spherical  triangle  ABC,  and  also  that  a  diedral 
angle  is  measured  by  its  plane  angle. 

There  is  then  a  correspondence  between  the  triedral 
angle  0-ABC  and  the  spherical  triangle  ABC :  the  six 
parts  of  the  triedral  angle  are  represented  by  the  corre- 
sponding six  parts  of  the  spherical  triangle,  and  all  the 
relations  among  the  parts  of  the  former  are  the  same  as 
the  relations  among  the  corresponding  parts  of  the  latter. 

184.  Fundamental  Definitions  and  Properties.  —  The  fol- 
lowing definitions  and  properties  are  from  Geometry,  Book 
VIII.  : 

In  every  spherical  triangle 

Each  side  is  less  than  the  sum  of  the  other  two. 

The  sum  of  the  three  sides  lies  between  0°  and  360°. 

The  sum  of  the  three  angles  lies  between  180°  and  540°. 

Each  angle  is  greater  than  the  difference  between  180° 
and  the  sum  of  the  other  two. 

If  two  sides  are  equal,  the  angles  opposite  them  are 
equal ;  and  conversely. 

If  two  sides  are  unequal,  the  greater  side  lies  opposite 
the  greater  angle  ;  and  conversely. 

The  perpendicular  from  the  vertex  to  the  base  of  an 
isosceles  triangle  bisects  both  the  vertical  angle  and  the 
base. 


DEFINITIONS  AND   PROPERTIES.  269 

The  axis  of  a  circle  is  the  diameter  of  the  sphere  perpen- 
dicular to  the  plane  of  the  circle.  The  poles  of  a  circle  are 
the  two  points  in  which  its'  axis  meets  the  surface  of  the 
sphere. 

One  spherical  triangle  is  called  the  polar  triangle  of  a 
second  spherical  triangle  when  the  sides  of  the  first  triangle 
have  their  poles  at  the  vertices  of  the  second. 

If  the  first  of  two  spherical  triangles  is  the  polar  triangle 
of  the  second,  then  the  second  is  the  polar  triangle  of  the 
first. 

Two  such  triangles  are  said  to  be  polar  with  respect  to 
each  other.  Thus : 

If  A'B'C'  is  the  polar  triangle  of 
ABC,  then  ABC  is  the  polar  triangle 
of  A'B'C'. 

In  two  polar  triangles,  each  angle 
of  one  is  measured  by  the  supplement 
of  the  corresponding  side  of  the  other. 

Thus : 

A  =  180°  -  a'}        B  =  180°  -  b',        C  =  180°  -  c', 
a  =180°  -A',         6  =  180°-B',         c  =  180°-C'. 

This  result  is  of  great  importance ;  for  if  any  general 
equation  be  established  between  the  sides  and  angles  of  a 
spherical  triangle,  it  holds  of  course  for  the  polar  triangle 
also.  Hence,  by  means  of  the  above  formulas,  any  theorem  of 
a  spherical  triangle  may  be  at  once  transformed  into  another 
theorem  by  substituting  for  each  side  and  angle  respectively 
the  supplements  of  its  opposite  angle  and  side. 

If  a  spherical  triangle  has  one  right  angle,  it  is  called  a 
right  triangle ;  if.it  has  two  right  angles,  it  is  called  a  bi- 
rectangular  triangle ;  and  if  it  has  three  right  angles,  it  is 
called  a  tri-rectangular  triangle.  If  it  has  one  side  equal  to 
a  quadrant,  it  is  called  a  quadrantal  triangle ;  and  if  it  has 
two  sides  equal  to  a  quadrant,  it  is  called  a  bi-quadrantal 
triangle. 


270 


SPHERICAL    TRIG  ON  OMETR  Y. 


XOTE.  —  It  is  shown  in  geometry  that  a  spherical  triangle  may,  in  general,  be 
constructed  when  any  three  of  its  six  parts  are  given  (not  excepting  the  case  in 
which  the  given  parts  are  the  three  angles).  In  spherical  trigonometry  we  investi- 
gate the  methods  by  which  the  unknown  parts  of  a  spherical  triangle  may  be  com- 
puted from  the  above  data. 

EXAMPLES. 

1.  In  the  spherical  triangle  whose  angles  are  A,  B,  C, 
prove 

B  +  C-A<7r (1) 

C  +  A  -  B  <  TT (2) 

A  +  B-C<7r (3) 

2.  If  C  is  a  right  angle,  prove 

A  +  B  <  ITT  (1),  and  A  -  B  <  £  (2). 

3.  The  angles  of  a  triangle  are  A,  45°,  and  120° ;  find  the 
maximum  value  of  A.  Ans.  A  <  105°. 

4.  The  angles  of  a  triangle  are  A,  30°,  and  150° ;  find  the 
maximum  value  of  A.  Ans.  A  <  60°. 

5.  The  angles  of  a  triangle  are  A,  20°,  and  110°;  find  the 
maximum  value  of  A.  Ans.  A  <  90°. 

6.  Any  side  of  a  triangle  is  greater  than  the  difference 
between  the  other  two. 


RIGHT  SPHERICAL  TRIANGLES. 

185.   Formulae  for   Right    Triangles.  —  Let    ABC    be    a 

spherical  triangle  in  which  C  is 
a  right  angle,  and  let  0  be  the 
centre  of  the  sphere ;  then  will 
OA,  OB,  OC  be  radii:  let  a,  6,  c 
denote  the  sides  of  the  triangle  O- 
opposite  the  angles  A,  B,  C,  re- 
spectively ;  then  a,  &,  and  c  are 
the  measures  of  the  angles  BOG, 
COA,  and  AOB. 


RIGHT  SPHERICAL   TRIANGLES.  271 

. 

From  any  point  D  in  OA  draw  DE  _L  to  £)&,  and  from  E 
draw  EF  _L  to  erf/Snd  join  DF.  Then  BE  is  JL  to  EF  (Geom. 
Art.  537).  Hence  (Geom.  Art.  507),  DF 

DF  is  JL  to  O^rf  .-.  Z  DFE  =  Z  §£  •     (Art-  183) 

OTP       OTT    OT?  v  ^ 

Now  —  —  —  -  — ;  that  is,  cos  fc  =  cos  a  cos  to  S     .     (1) 
OD      OE    OD 

^=il'^;  thatis'sin6^sillBsinc-  •  (2) 

Interchanging  a's  and  &'s,  sin  a  =  sin  A  sin  c  .  .  (3) 

§|  =  ||  •  ^f;  that  is,  tan  o=  cos  B  tan  c.  .  (4) 

Interchanging  a's  and  6's,  tan  b  =  cos  A  tan  c  .  .  (5) 

5E  =  55.5Z;  that  is,  tan  6  =  tan  B  sin  a.  .  (6) 

Interchanging  a's  and  6's,        tan  a  =  tan  A  sin  ?> .     .     (7) 
Multiply  (6)  and  (7)  together,  and  we  get 

tan  A  tan  B  =  -  -  =  -J— ,  by  (1) 

cos  a  cos  b      cos  c 

.-.  cos  c  =  cot  A  cot  B ($) 

Multiply  crosswise  (3)  and  (4),  and  we  get 
sin  a  cos  B  tan  c  =  tan  a  sin  A  sin  c. 

-r,      sin  A  cos  c  A         *  <u     /«t\ 

.-.  cosB  =  —         =  sin  A  cos  6,  by  (1)    ...     (9) 

cos  a 

Interchanging  a's  and  6's, 

cos  A  =  sin  B  cos  a (10) 

Sch.  By  these  ten  formulae,  every  case  of  right  triangles 
can  be  solved  j  for  every  one  of  these  ten  formulae  is  a  dis- 
tinct combination,  involving  three  out  of  the  five  quantities, 
a,  b,  c,  A,  B,  and  there  can  be  but  ten  combinations  in  all. 
Hence,  any  two  of  the  five  quantities  being  given  and  a 
third  required,  that  third  quantity  may  be  determined  by 
some  one  of  the  above  ten  formulae. 


272  SPHERICAL    TRIGONOMETRY. 

186.  Napier's  Rules.  — The  ten  preceding  formulae,  which 
may  be  found  difficult  to  remember,  have  been  included 
under  two  simple  rules,  called  after  their  inventor,  Napier's 
Hides  of  the  Circular  Parts. 

Let  ABC  be  a  right  spherical  triangle.     Omit  the  right 
angle  C.     Then  the  two  sides  a 
and   b,   which  include  the  right 
angle,    the    complement    of    the 
hypotenuse   c,    and   the    comple- 
ments of   the  oblique  angles  A 
and  B,  are  called  the  circular  parts 
of  the  triangle.     Thus,  there  are 
five  circular  parts,  arranged  in  the  figure  in  the  following 
order :  a,  b}  co.  A,  co.  c,  co.  B. 

Any  one  of  these  five  parts  may  be  selected  and  called 
the  middle  part;  then  the  two  parts  next  to  it  are  called 
adjacent  parts,  and  the  remaining  two  parts  are  called  oppo- 
site parts.  Thus,  if  co.  A  is  selected  as  the  middle  part, 
then  b  and  co.  c  are  the  adjacent  parts,  and  a  and  co.  B  are 
the  opposite  parts. 

Then  Napier's  Rules  are :  - 

(1)  TJie  sine  of  the  middle  part  equals  the  product  of  the 
tangents  of  the  adjacent  parts. 

(2)  The  sine  of  the  middle  part  equals  the  product  of  the 
cosines  of  the  opposite  parts. 

NOTE  1.  —  It  will  assist  the  student  in  remembering  these  rules  to  notice  the 
occurrence  of  the  vowel  i  in  sine  and  middle,  of  the  vowel  a  in  tangent  and  adjacent, 
and  of  the  vowel  o  in  cosine  and  opposite. 

Napier's  Rules*  may  be  made  evident  by  taking  in  detail  each  of  the  five  parts  as 
middle  part,  and  comparing  the  equations  thus  found  with  the  formulae  of  Art.  18o. 

Thus,  let  co.  c  be  the  middle  part.    The  rules  give 

sin  (co.  c)  =  tan  (co.  A)  tan  (co.  B);  /.  cos  c  =  cot  A  cot  B (8) 

sin(co.  c)=cos  a  cos  6;  /.  cos  c  =cos«  cos  6 (1) 

co.  B  the  middle  part. 

sin(co.  B)=  tan  a  tan(co.  c) ;  .'.  cos  B  =  tan  a  cot  c (4) 

sin(co.  B)  =  cos  6  cos(co.  A) ;  .•.  cos  B  =  cos  6  sin  A (9) 

*  While  some  find  these  rules  to  be  useful  aids  to  the  memory,  others  question 
their  utility. 


THE  SPECIES   OF  THE  PARTS.  273 

a  the  middle  part. 

sin  a  =  tan  b  tan (co.  B) ;  .-.  sin  a  =tan  6  cot  B (6) 

sin  a  =  cos(co.  A)cos(co.  c) ;     .*.  sin  a  =  sin  A  sine (3) 

b  the  middle  part. 

sin  6  =  tan  a  tan(co.  A) ;  /.  sin  6  =  tan  a  cot  A (7) 

sin  &  =  cos(co.  c)cos(co.  B) ;      /.  sin  6  =  sine  sin  B (2) 

co.  A  the  middle  part. 

sin(co.  A)=  tan  6  tan(co.  c) ;     .•.  cos  A=  tan  b  cot  c (5) 

sin(co.  A)=  COB  a  cos(co.  B) ;    .'.  cos  A=  cos  a  Bin  B (10) 

NOTE  2. —  In  applying  these  rules  it  is  not  necessary  to  use  the  notation  co.  c, 
co.  A,  co.  B,  since  we  may  write  at  once  cos  c  for  sin  (co.  c),  etc. 

187.  The  Species  of  the  Parts.  —  If  two  parts  of  a  spheri- 
cal triangle  are  either  both  less  than  90°  or  both  greater  than 
90°,  they  are  said  to  be  of  the  same  species.  But  if  one  part 
is  less  than  90°  and  the  other  part  is  greater  than  90°,  they 
are  of  different  species. 

In  order  to  determine  whether  the  required  parts  are  less 
or  greater  than  90°,  it  will  be  necessary  carefully  to  observe 
their  algebraic  signs.  If  the  required  part  is  determined 
by  means  of  its  cosine,  tangent,  or  cotangent,  the  alge- 
braic sign  of  the  result  will  show  whether  it  is  less  or 
greater  than  90°.  But  when  a  required  part  is  found  in 
terms  of  its  sine,  it  will  be  ambiguous,  since  the  sines  are 
positive  in  both  the  first  and  second  quadrants.  This 
ambiguity,  however,  may  generally  be  removed  by  either 
of  the  following  principles  : 

(1)  In  a  right  spherical  triangle,  either  of  the  sides  con- 
taining the  right  triangle  is  of  the  same  species  as  the  opposite 


(2)  The  three  sides  of  a  right  spherical  triangle  (omitting 
bi-rectangular  or  tri-rectangular  triangles)  are  either  all 
acute,  or  else  one  is  acute  and  the  other  two  obtuse. 

The  first  follows  from  the  equation 
cos  A  =  cos  a  sin  B, 


t  • 

274  SPHERICAL    TRIGONOMETRY. 

in  which,  since  sinB  is  always  positive  (B  <  180°),  cos  A 
and  cos  a  must  have  the  same  sign;  i.e.,  A  and  a  must  be 
either  both  <  or  both  >  90°. 

The  second  follows  from  the  equation 

cos  c  =  cos  a  cos  b. 

188.  Ambiguous  Solution.  —  When  the  given  parts  of  a 
right  triangle  are  a  side  and  its  opposite  angle,  the  triangle 
cannot  be  determined. 

For  two  right  spherical  triangles  ABC,  A'BC,  right 
angled  at  C,  may  always  be 
found,  having  the  angles  A 
and  A'  equal,  and  BC,  the 
side  opposite  these  angles, 
the  same  in  both  triangles, 

but  the  remaining  sides,  AB,  AC,  and  the  remaining  angle 
ABC  of  the  one  triangle  are  the  supplements  of  the  re- 
maining sides  A'B,  A'C,  and  the  remaining  angle  A'BC  of 
the  other  triangle.  It  is  therefore  ambiguous  whether 
ABC  or  A'BC  be  the  triangle  required. 

This   ambiguity  will   also   be   found   to   exist,  if  it   be 
attempted  to  determine  the  triangle  by  the  equation 

sin  6  =  tan  a  cot  A, 

since  it  cannot  be  determined  from  this  equation  whether 
the  side  AC  is  to  be  taken  or  its  supplement  A'C. 

189.  Quadrantal  Triangles,  —  The.  polar   triangle   of   a 
right  triangle  has  one  side  a  quadrant,  and  is  therefore 
a  quadrantal  triangle  (Art.  184).     The  formulae  for  quad- 
rantal    triangles    may   be    obtained  by   applying  the   ten 
formulae  of  Art.  185  to  the  polar  triangle.     They  are  as 
follows,  c  being  the  quadrantal  side : 

cos  C  =  —  cos  A  cos  B (1) 

sin  B  =  sin  b  sin  C (2) 


EXAMPLES.  275 

sin  A  =  sin  a  sin  C (3) 

cos  b  =  —  tan  A  cot  C (4) 

cos  a  =  —  tan  B  cot  C (5) 

sin  A  =  tan  B  cot  b (6) 

sin  B  =  tan  A  cot  a (7) 

cos  C  =  —  cot  a  cot  b (8) 

c6s  6  =  cos  B  sin  a (9) 

cos  a  =  cos  A  sin  b (10) 

EXAMPLES. 

In  the  right  triangle  ABC  in  which  the  angle  C  is  the 
right  angle,  prove  the  following  relations : 

1.  sin2  a  -f  sin2  b  —  sin2  c  =  sin2 a  sin2  b. 

2.  cos2  A  sin2  c  =  sin2  c  —  sin2  a. 

3.  sin2  A  cos2  c  =  sin2  A  —  sin2  a. 

4.  sin2  A  cos2  b  sin2c  =  sin2c  —  sin2&. 

5.  2  cos  c  =  cos  (a  +  &)  +  cos  (a  —  b). 

6.  tan  i  (c  +  a)  tan  1  (c  —  a)  =  tan2 1 6. 

7.  sin2 -=  sin2 -cos2 -4- cos2 -sin2-- 

2  2222 

8.  sin  (c- 6)  =  tan2  —  sin(c  +  6). 

2i 

9.  If  b  =  c  =  ^,  prove  cos  a  =  cos  A. 

10.  If  a  =  b  =  c}  prove  sec  A  =  1  -f  sec  a. 

11.  If  c  <  90°,  show  that  a  and  6  are  of  the  same  species. 

12.  If  c  >  90°,  a  and  b  are  of  different  species. 

13.  A  side  and  the  hypotenuse  are  of  the  same  or  oppo- 
site species,  according  as  the  included  angle  <;  or  >-• 


276  SPHERICAL   TRIGONOMETRY. 

OBLIQUE  SPHERICAL  TRIANGLES. 

190.  Law  of  Sines.  —  In  any  spherical  triangle  the  sines 
of  the  sides  are  proportional  to  the  sines  of  the  opposite  angles. 

Let  ABC  be  a  spherical  triangle,  0  the  centre  of  the 
sphere ;  and  let  a,  6,  c  denote  the 
sides  of  the  triangle  opposite  the 
angles  A,  B,  C,  respectively.  Then 
a,  6,  and  c  are  the  measures  of  the 
angles  BOC,  CO  A,  and  AOB. 

From  any  point  D  in  OA  draw 
DG  _L  to  the  plane  BOC,  and  from 
G  draw  GE,  GF  J_  to  OB,  OC. 
Join  DE,  DF,  and  GO.  Then  DG 
is  J_  to  GE,  GF,  and  GO  (Geom.  Art.  487).  Hence,  DE  is 
_L  to  OB,  and  DF  _L  to  OC  (Geom.  Art.  507). 

.-.  Z  DEG  =F  Z  B,  and  Z  DFG  =  Z  C  .     .   (Art.  183) 
In  the  right  plane  triangles  DGE,  DGF,  ODE,  ODF, 
DG  =  DE  sin  B  =  OD  sin  DOE  sin  B  =  OD  sin  c  sin  B, 
DG  =  DF  sin  C  =  OD  sin  DOF  sin  C  =  OD  sin  b  sin  C. 

.-.  sin  c  sin  B  =  sin  b  sin  C  ; 
or  sin  6  :  sin  c  :  :  sin  B  :  sin  C. 

Similarly,  it  may  be  shown  that 

sin  a  :  sin  c  : :  sin  A  :  sin  C. 

sin  a  _  sin  b  _  sin  c 
sin  A      sin  B      sin  C 

NOTE.  — The  common  value  of  these  three  ratios  is  called  the  modulus  of  the 
spherical  triangle. 

JSch.  In  the  figure,  B,  C,  6,  c  are  each  less  than  a  right 
angle ;  but  it  will  be  found  011  examination  that  the  proof 
will  hold  when  the  figure  is  modified  to  meet  any  case 
which  can  occur.  For  example,  if  B  alone  is  greater  than 


LAW  OF  COSINES.  277 

90°,  the  point  G  will  fall  outside  of  OB  instead  of  between 
OB  and  OC.  Then  DEG  will  be  the  supplement  of  B,  and 
thus  we  shall  still  have  sin  DEG  =  sin  B. 

191.  Law  of  Cosines.  —  In  any  spherical  triangle,  the 
cosine  of  each  side  is  equal  to  the  product  of  the  cosines  of  the 
other  two  sides,  plus  the  product  of  the  sines  of  those  sides 
into  the  cosine  of  their  included  angle. 

Let  ABC  be  a  spherical  triangle,  0  the  centre  of  the 
sphere,  and  a,  &,  c  the  sides  of 

the  triangle  opposite  the  angles  D^-^1\ 

A,  B,  C,  respectively.     Then  ^^n       IV 

a  =  ZBOC,  0<^  /%    /    )c 

b  =  Z  COA,  ^"^l^X 

c  =  Z  AOB.  B 

From  any  point  D  in  OA  draw,  in  the  planes  AOB, 
AOC,  respectively,  the  lines  DE,  DF  1_  to  0  A.  Then 

Z  EDF  =  Z  A   .......     (Art.  183) 

Join  EF;  then  in  the  plane  triangles  EOF,  EDF,  we 
have 

OE2  +  OF2-20E-OFcosEOF  .     .     (1) 

DE2  +  DF2-2DE.DFcosEDF  .     .     (2) 
also  in  the  right  triangles  EOD,  FOD,  we  have 

OE2  -  DE2  =  OD2,  and  OF2-DF2  =  OD2   .     (3) 

Subtracting  (2)  from  (1),  and  reducing  by  (3),  and 
transposing,  we  get 

2  OE  .  OF  cos  EOF  =  2  OD2  +  2  DE  •  DF  cos  EDF. 


or  cos  a  =  cos  b  cose  +  sin  b  sine  cos  A     (4) 


278  SPHERICAL   TRIGONOMETRY. 

By  treating  the  other  edges  in  order  in  the  same  way, 
or  by  advancing  letters  (see  Note,  Art.  9G)  we  get 

cos  b  =  cos  c  cos  a  4  sin  c  sin  a  cos  B      .     .     (5) 
cos  c  =  cos  a  cos  b  4  sin  a  sin  &  cos  C      .     .     (6) 

Sch.  Formula  (4)  has  been  proved  only  for  the  case  in 
which  the  sides  b  and  c  are  less  than  quadrants;  but  it 
may  be  shown  to  be  true  when  these  sides  are  not  less  than 
quadrants,  as  follows : 

(1)  Suppose  c  is  greater  — — 

than  90°.     Produce  B  A,  BC 

to    meet    m    B',    and    put 
AB'=c',  CB'=a'. 

Then,  from  the  triangle  AB'C,  we  have  by  (4) 
cos  a'  =  cos  b  cos  cf  4  sin  b  sin  cf  cos  B'AC, 
or  COS(TT— a)  =  cos  b  cos(?r— c)  4  sin  b  sin(?r  —  C)COS(TT  —A). 
.  •.  cos  a  =  cos  b  cos  c  -f-  sin  b  sin  c  cos  A. 

(2)  Suppose  both  b  and  c  to 
be  greater  than  90°.    Produce 
AB,  AC  to  meet  in  A',  and  put 
A'B  =  c',  A'C  =  b'. 

Then,  from  the  triangle  A'BC,  we  have  by  (4) 
cos  a  =  cos  b'  cos  c'  4-  sin  b'  sin  c'  cos  Ar; 
but  b1  =  TT  -  6,  c'  =  TT  -  c,  A'  =  A. 

.-.  cos  a  =  cos  b  cos  c  4  sin  b  sin  c  cos  A. 
The  triangle  AB'C  is  called  the  colunar  triangle  of  ABC. 

192.  Relation  between  a  Side  and  the  Three  Angles.  — 
In  any  spherical  triangle  ABC, 

cos  A  =  —  cos  B  cos  C  4  sin  B  sin  C  cos  a. 


RELATION  BETWEEN  SIDE  AND  ANGLES.      279 

Let  A'B'C'  be  the  polar  triangle  of  ABC,  and  denote  its 
angles  and  sides  by  A',  B',  C',  a',  b',  c' ;  then  we  have  by 
(4)  of  Art.  191 

cos  a'  =  cos  b'  cos  c'  +  sin  b'  sin  c'  cos  A' ; 
but       a'  =  TT  -  A,  b'  =  TT  -  B,  c'  =  TT  -  C,  etc.      .  (Art.  184) 
Hence,  substituting,  we  get 

cos  A  =  —  cos  B  cos  C  +  sin  B  sin  C  cos  a    .     .     .     .     (1) 
Similarly, 

cos  B  =  —  cos  C  cos  A  -j-  sin  C  sin  A  cos  &   ....     (2) 

cos  C  =  —  cos  A  cos  B  -f-  sin  A  sin  B  cos  c  .     .     .     .     (3) 

Hem.  —  This  process  is  called  "  applying  the  formula  to  the  polar  triangle."  By 
means  of  the  polar  triangle,  any  formula  of  a  spherical  triangle  may  be  immediately 
transformed  into  another,  in  which  angles  take  the  place  of  sides,  and  sides  of  angles. 

193.  To  show  that  in  a  spherical  triangle  ABC, 

cot  a  sin  b  =  cot  A  sin  C  +  cos  C  cos  b. 

Multiply  (6)  of  Art.  191  by  cos  b,  and  substitute  the 
result  in  (4)  of  Art.  191,  and  we  get 

cos  a  =  cos  a  cos2  b  +  sin  a  sin  b  cos  b  cos  C  -f-  sin  b  sin  c  cos  A. 
Transpose  cos  a  cos2  b,  and  divide  by  sin  a  sin  b ;  thus, 

cot  a  sin  b  =  cosb  cos  C  +  Sm  °  C°S  A 

sin  a 

=  cos  b  cos  C  +  cot  A  sin  C    .    (by  Art.  190) 

By  interchanging  the  letters,  we  obtain  five  other  formulae 
like  the  preceding  one.  The  six  formulae  are  as  follows  : 

cot  a  sin  b  =  cot  A  sin  C  +  cos  C  cos  b  .     .     .     .  (1) 

cot  a  sin  c  =  cot  A  sin  B  +  cos  B  cos  c  .     .     .     .  (2) 

cot  b  sin  a  =  cot  B  sin  C  -f-  cos  C  cos  a  .     .     .     .  (3) 

cot  b  sin  c  =  cot  B  sin  A  -f-  cos  A  cos  c  ....  (4) 

cot  c  sin  a  =  cot  C  sin  B  +  cosB  cos  a  ....  (5) 

cot  c  sin  b  =  cot  C  sin  A  +  cos  A  cos  b  .     .     .     .  (6) 


280  SPHERICAL   TRIGONOMETRY. 

EXAMPLES. 

1.  If  a,  b,  c  be  the  sides  of  a  spherical  triangle,  a',  b',  c' 
the  sides  of  its  polar  triangle,  prove 

sin  a  :  sin  b  :  sin  c  =  sin  a' :  sin  6' :  sin  c'. 

2.  If  the   bisector  AD   of  the  angle  A  of  a   spherical 
triangle  divide  the  side  BC  into   the   segments   CD  =  b', 
BD  =  c',  prove 

sin  b  :  sin  c  =  sin  6' :  sin  c'. 

3.  If  D  be  any  point  of  the  side  BC,  prove  that 

cot  AB  sin  DAC  +  cot  AC  sin  DAB  =  cot  AD  sin  BAG. 
cot  ABC  sin  DC  +  cot  ACB  sin  BD  =  cot  ADB  sin  BC. 

4.  If  a,  /?,  y  be  the  perpendiculars  of  a  triangle,  prove  that 

sin  a  sin  a  =  sin  b  sin  (3  —  sin  c  sin  y. 

5.  In  Ex.  4  prove  that 


sin  a  cos  a  =  Vcos2  b  +  cos2  c  —  2  cos  a  cos  b  cos  c. 

194.  Useful  Formulae.  —  Several  other  groups  of  useful 
formulae  are  easily  obtained  from  those  of  Art.  191 ;  the 
following  are  left  as  exercises  for  the  student : 

sin  a  cos  B  =  cos  b  sin  c  —  sin  b  cose  cos  A  .  .  (1) 

sin  a  cos  C  =  sin  b  cos  c  —  cos  b  sin  c  cos  A  .  .  (2) 

sin  b  cos  A  =  cos  a  sin  c  —  sin  a  cose  cos  B  .  .  (3) 

sin  b  cos  C  =  sin  a  cos  c  —  cos  a  sin  c  cos  B  .  .  (4) 

sin  c  cos  A  =  cos  a  sin  b  —  sin  a  cos  b  cos  C  .  .  (5) 

sin  c  cos  B  =  sin  a  cos  b  —  cos  a  sin  b  cos  C  .  .  (6) 


FORMULA  FOR    THE  HALF  ANGLES.  281 

Applying  these  six  formulae  to  the  polar  triangle,  we 
obtain  the  following  six  : 

sin  A  cos  b  =  cos  B  sin  C  +  sin  B  cos  C  cos  a  .  .  (7) 

sin  A  cos  c  =  sin  B  cos  C  +  cos  B  sin  C  cos  a  .  .  (8) 

sin  B  cos  a  =  cos  A  sin  C  +  sin  A  cos  C  cos  b  .  .  (9) 

sin  B  cose  =  sin  A  cos  C  +  cos  A  sin  C  cos  b  .  .  (10) 

sin  C  cos  a  =  cos  A  sin  B  -f-  sin  A  cos  B  cos  c  .  .  (11) 

sin  C  cos  b  =  sin  A  cos  B  -f-  cos  A  sin  B  cos  c  .  .  (12) 

195.  Formulae  for  the  Half  Angles.  —  To  express  the 
sine,  cosine,  and  tangent  of  half  an  angle  of  a  spherical 
triangle  in  terms  of  the  sides. 

I.   By  (4)  of  Art.  191  we  have 

cos  A  =  cos  a-  cos  fr  cose  =  ±  _  2  sin,  A  (Art  49) 
sin  b  sin  c  2 

o   •  2-A.__-i      cos  a  —  cos  6  cos  c 
2  sin  b  sin  c 

_  cos  (b  —  c)  —  cos  a 
sin  b  sin  c 


/Art  45) 


Let  2s  =  a  +  6  +  c;  so  that  s  is  half  the  sum  of  the  sides 
of  the  triangle  ;  then 

a  +  b  —  c  =  2(s  —  c),  and  a  —  b  +  c  =  2(s  —  b). 

•  2  A      sin  (s  —  b)  sin  (s  —  c) 
.'.  sin  —  =  --  *  -  !  —  ;  —     -  • 
sin  b  sin  c 


2  sin  b  sin  c 


A         /sin  (s  —  b}  sin  (s  —  c)  /^  ^ 

.-.    8in-=-J  '    .    .     •     (l) 

2       \  sin  b  sine 


282  SPHERICAL    TRIGONOMETRY. 

Advancing  letters, 

*>^»<-««»<-« 


sn  c  sn  a 


sinC       /!i!L(l^)si"( 
2       \  sin  a  sin  6 


II.  2cos2     =  l  +  cosA       .....     (Art.  49) 

2 

-,   .  cos  a  —  cos  b  cos  c 

s=  i-j  --  ;  —  -  —  :  - 

sin  o  sin  c 

_  cos  a  —  cos  (b  +  c) 
sin  6  sin  c 

2A_  sin4-(a  4-  b  +  c)  sin|(6  4-  c  —  a) 

.*.    COS   -•--  —  -  -  ;  —  -  —  :  —  - 

2  sin  b  sin  c 

_  sins  sin(s  —  a) 
sin  b  sine 


sin  6  sin  c 
Advancing  letters, 


C=    Isins 

2       \      si 


sin  a  sin  b 
III.    By  division,  we  obtain 


2       \      sin  s  sin  (s  —  a) 

•-&   .    .    .     (8) 


2       \       sin  s  sin  (N  — 


tan  —  =    /sin(g~q;)  «i"(-'*  — 
2       if       sins  sin  (s  — c) 


FORMULA  FOR   THE  HALF  ANGLES.  283 

Sch.  The  positive  sign  must  be  given  to  the  radicals  in 
each  case  in  this  article,  because  JA,  J-B,  ^C  are  each  less 
than  90°. 

Cor.  1. 


tan      tan       —  s 

in  (s  —  c) 

nch 

tcUl         Ld,lJ          — 

sins 
in(s  —  a) 

.      ^-LV; 

ni\ 

tan  2  tan  - 

•j-qi-j        till         — 

sins 

C12^ 

belli        belli 

sins 

•  {-*-*) 

Cor.  2.    Since  sin  A  =  2  sin  -cos  A, 


in  A    _  2Vsins  sin(s  —  a)  sin(s  —  b)  sin(.s-  —  c)          ,|ox 
sin 6  sine 

=  ^~        (14) 

sin  o  sin  c 

where  n2  =  sins  sin(s  —  a)  sin(s  —  b)  sin(s  —  c). 


1.   Prove  sin2A  = 


EXAMPLES. 

1 — cos2a  —  cos2  6  —  cos2e  -f 2  cos  a  cos  b  cos  c 
sin"  b  siir'c 


sin2  6  sin2c' 
where  4  ir  =  1  —  cos2  a  —  cos2  6  —  cos2  c  -f  2  cos  a  cos  b  cos  c. 

2.  Prove  cose  =  cos  (a  +  b)  sin2 — f-  cos  (a  —  b)  cos2- — 

.    A   .    B    .    C     sin(s— a)  sin  (s-^fr)sin(s— e) 

3.  Prove  sm  —  sin  —  sin  — = —  — . 

222  sin  a  sin  b  sine 

4.  prove   cos  A  +  cos  B  .=  sin  (a  +  b) 

1  — cosC  sine 

5.  Prove  SC°sA +  COsB  sin  (a -6)sinc  =  0. 

1  —  cos  C 

6.  Prove   CQS  A  -  cos  B  =  sin  (a~b\ 

1  +  cosC  sine 


284  SPHERICAL    TRIGONOMETRY. 

196.  Formulae  for  the  Half  Sides.  —  To  express  the  sine, 
cosine,  and  tangent  of  half  a  side  of  a  spherical  triangle  in 
terms  of  the  angles. 

By  (1)  of  Art.  192,  we  have 


cos  a  =  +  cos  B  cos  C  =  ±  _  2  gin2a  A 

sin  B  sin  C  2 

2  sin2a  =      cos  A  +  cos  (B  +  C)  . 
2  sin  B  sin  C 


(Art 


2  sin  B  sin  C 

Let   2S  =  A  +  B  +  C;  then  B  +  C  -  A  =  2  (S  -  A). 

Proceeding  in  the  same  way  as  in  Art.  195,  we  find  the 
following  expressions  for  the  sides,  in  terras  of  the  three 
angles : 

.    a_     /     cos  S  cos  (8  —  A)  ,^ 

m2~  \  sin  B  sin  C 

cos  S  cos  (S  —  B)  ,£\ 

2      \  sin  C  sin  A 


fcW 


sm^,_cosScos(S-C) 
sin  A  sin  ±5 


cos- 


a        /cos  (S  -  B)  cos  (S  -  C)  m 

2      \  sinBsinC 


-v 


cos  6  _  .  /cos(S-C)cos(S-A) 


2      \  sin  C  sin  A 


ogc_      cos  (S  -  A)  cos  (S  -  B) 
2      Af  sin  A  sin  B 

t     °L—     /3       cosS  cos  (S  —  A) 
m2  "  \  ~  cos  (S  -  B)  cos  (S  -  C) 


FORMULAE  FOE    THE  HALF  SIDES.  285 


t     5=     /_       cos  S  cos  (S  -  B) 
2      \      cosS-CcosS 


cos(S-C)cos(S-A) 


tan-=     /—        cos  S  cos  (S  -  C) 
2      \      cosfS-AUosrS- 


Sch.  1.  These  formulae  may  also  be  obtained  immediately 
from  those  of  Art.  195  by  means  of  the  polar  triangle. 

Sch.  2.  The  positive  sign  must  be  given1  to  the  above 
radicals,  because  ^,  -,  ^,  are  each  less  than  90°. 

a    £  £t 

Sch.  3.  These  values  of  the  sines,  cosines,  and  tangents 
of  the  half  sides  are  always  real. 

For  S  is  >  90°  and  <  270°  (Art.  184),  so  that  cos  S  is 
always  negative. 

Also,  in  the  polar  triangle,  any  side  is  less  than  the  sum 
of  the  other  two  (Art.  184). 

.'.     7T  —  A<7T  —  B  +  7T—  C. 

...  B  +  C-A<7r. 

.-.  cos  (S  —  A)  is  positive,. 
Similarly,    cos  (S  —  B)  and  cos  (S  —  C)  are  positive. 


Cor.    Since  sin  a  =  2  sin  -  cos  -, 

_2V—  cosScos(S  —  A)cos(S  —  B)cos(S— C) 

.  *.    Sill  Ct»  — ~ — ~ 

sin  B  sin  C 

2N 

;) 


sinB  sinC 


where  N=  V-  cos  S  cos  (S  -  A)  cos  (S  -  B)  cos  (S  -  C). 


286  SPHEEICAL    TRIGONOMETRY. 

EXAMPLES. 

1.  Prove  cosC=-cos(A  +  B)cos2--cos(A-B)sin2-. 

0    -r,  .    a   .    b    •    c  —  N  cos  S 

2.  Prove  sin  -  sin  -  sin  -  =  —  — , 

222      sin  A  sin  B  sin  C 

where  N  «=  V  —  cos  S  cos  (S  —  A)  cos  (S  —  B)  cos  (S  —  C). 

197.  Napier's  Analogies. 

T   i  sin  A      sin  B  ,  A    ,    -4<\r\\  /-« \ 

Let  m  =  —    —  =—    -  (Art.  190)  (1) 

sin  a       sin  6 

sin  A  +  sin  B  ,  . ,     , 
=  -r-  —-(Algebra)  (2) 

sin  a  4-  sin  o 

or  =  sin  A- sin  B  (3) 

sin  a  ~  sin  6 

cos  A  4-  cos  B  cos  C  =  sin  B  sin  C  cos  a   (Art.  192) 
=  m  sin  C  sin  b  cos  a,  by  (1 )  (4) 
and  cos  B  +  cos  C  cos  A  =  sin  C  sin  A  cos  b 

=  m  sin  C  sin  a  cos  b  .     .     (5) 

.  •.   (cos  A  +  cos  B)  (1  4-  cos  C)  =  m  sin  C  sin  (a  4-  b) ,          (6) 

from  (4)  and  (5) 
Dividing  (2)  by  (6), 

sin  A  4-  sin  B  _  sin  a  4-  sin  b    1  4-  cos  C 
cos  A  4-  cos  B        sin  (a.  4-  b)         sin  C 

...  tani(A  +  B)  =  costitt-;icc4  (7) 

cos  4  (Oi  4-  b)        2, 

(Arts.  45,  46,  and  49) 

Similarly,       tan^  (A  -  B)  =  s|n^a  ~  ^  cotg  .     .     (8) 

sin  \  (a  4-  o)        * 


DEL  AMBERS  ANALOGIES.  287 

Writing  TT  —  A  for  a,  etc.,  by  Art.  184,  we  obtain  from 
(7)  and  (8) 


Sch.  The  formulae  (7),  (8),  (9),  (10)  are  known  as 
Napier1  s  Analogies,  after  their  discoverer.  The  last  two 
may  be  proved  without  the  polar  triangle  by  starting  with 
the  formulae  of  Art.  191. 

Cor.  In  any  spherical  triangle  whose  parts  are  positive, 
and  less  than  180°,  the  half-sum  of  any  two  sides  and  the  half- 
sum  of  their  opposite  angles  are  of  the  same  species. 

C 
For,  since  cos  £  (a  —  b)  and  cot  -  are  necessarily  positive, 

£ 

therefore  by   (7)  tan  £  (A  -f-  B)  and  cos  J  (a  +  b)  are  both 
positive  or  both  negative. 

.-.  £  (  A  +  B)  and  £  (a  -f  b)  are  both  >  or  both  <  or  both 
=  90°. 

198.   Delambre's  (or  Gauss's  )  Analogies. 


=  sin  —  cos  — j-  cos  —  sin  — 

a  ~&  a  £    ' 

Vsin  (s  —  b)  sin  (s  —  c)     /sin  s  sin  (s  —  6) 
sin  b  sin  c  Ai      sin  c  sin  a 

_i_     /sins sin  (.s  — a)     /sin  (s  —  c)  sin  (s  — ^) 
A/      sin  b  sin  c       Al  sin  c  sin  a 


_  sin  (s  —  6)  -j-  sin  (s  —  a)     /sin  .9  sin  (s  —  c) 
sin  c  \      sin  a  sin  b 


a  —  b)        C  .K       ,  .,nKN 

cos       .....     (Arts.  45  and  195) 


cos- 


288  SPHERICAL    TRIGONOMETRY. 


.-.  sin  £  (A  +  B)  cos  |  =  cos  |  (a- 6)  cos-      .     .     .      (1) 

+j  £ 

Similarly,  we  obtain  the  following  three  equations  : 

sini(A-B)sin-  =  sini(a-&)cos-      ...     (2) 

2  2 

cos  i  (A  +  B)  cos  £  =  cos  \  (a  +  6)  sin  -     .     .     .     (3) 

2  2 

cosf  (A  — B)  sin- =  sin  j- (a +  6)  sin.  -     ...     .     (4) 

^  — 

Sch.  1.  When  the  sides  and  angles  are  all  less  than  180°, 
both  members  of  these  equations  are  positive. 

Sch.  2.  Napier's  analogies  may  be  obtained  from  De- 
lambre's  by  division. 

NOTE. — Delambre's  analogies  were  discovered  by  him  in  1807,  and  published  in 
the  Connaissance  des  Temps  for  1809,  p.  443.  They  were  subsequently  discovered 
independently  by  Gauss,  and  published  by  him,  and  are  sometimes  improperly 
called  Gauss's  equations.  Both  systems  may  be  proved  geometrically.  The 
geometric  proof  is  the  one  originally  given  by  Delambre.  It  was  rediscovered  by 
Professor  Crofton  in  1869,  and  published  in  the  Proceedings  of  the  London  Mathe- 
matical Society,  Vol.  III.  [Casey's  Trigonometry,  p.  41]. 


EXAMPLES. 

In  the  right  triangle  ABC,  in  which  C  is  the  right  angle, 
prove  the  following  relations  in  Exs.  1-45  : 

1.  sin2acos2&  =  sin  (c -f- 6)  sin  (c  —  &). 

2.  tan2  a  :  tan2  b  =  sin2c  —  sin2  b  :  sin2c  —  sin2tt. 

3.  cos2acos2B  =  sin2 A  —  sin2a. 

4.  cos2  A  +  cos2  c  =  cos2  A  cos2  c  +  cos2  a. 

5.  sin2  A  —  cos2  B  =  sin2  a  sin2  B. 

6.  If  one  of  the  sides  of  a  right  triangle  be  equal  to  the 
opposite  angle,  the  remaining  parts  are  each  equal  to  90°. 


EXAMPLES.  289 

7.  If  the  angle  A  of  a  right  triangle  be  acute,  show  that 
the  difference  of  the  sides  which  contain  it  is  less  than  90°. 

B      sin  (s  —  a) 

8.  Prove  tan  -  =  —  — '-> 

2  sins 

9.  Prove   (1)  2  w  =  sin  a  sin  6;   (2)  2N  =  sinasinB. 

10.  Prove     sin2 a  sin2  b  =  sin2 a  -f  sin2 b  —  sin2 c. 

11.  Prove  tan2 


2      sin  (c  +  6) 

12.  Prove  2sin2^=  sin2£(a  +  b)  +  sin2£(a-  b). 

13.  In  a  spherical  triangle,  if  c  =  90°,  prove  that 

tan  a  tan  b  +  sec  C  =  0. 

14.  In  a  spherical  triangle,  if  c  =  90°,  prove  that 

sin2p  =  cot  0  cot  <£, 

where  p  is  the  perpendicular  on  c,  and  0  and  <£  are  the  seg- 
ments of  the  vertical  angle. 

15.  Show  that  the  ratio  of  the  cosines  of  the  segments 
of  the  base  made  by  the  perpendicular  from  the  vertex  is 
equal  to  the  ratio  of  the  cosines  of  the  sides. 

16.  If  B  be  the  bisector  of  the  hypotenuse,  show  that 

sin'S  =  ^°  +  sin26 


4cos2- 


17.  Prove  tan  S  =  cot  -  cot  -. 

2       2 

18.  Construct  a  triangle,  being  given  the  hypotenuse  and 
(1)  the  sum  of  the  base  angles,  and  (2)  the  difference  of 
the  base  angles. 

19.  Given  the  hypotenuse  and  the  sum  or  difference  of 
the  sides  :  construct  the  triangle. 


290  SPHERICAL    TRIGONOMETRY. 

20.  Given  the  sum  of  the   sides  a  and  b,  and  the  sum 
of  the  base  angles  :  solve  the  triangle. 

21.  Show  that  sin^=^sin  c  +  sin  «  +  Vsin  c -sii[^ 

2  2Vsinc 


2  cos  ft  sin  c 
23.   cosA 


2  cos  b  sin  c 
24.    sin  (a  -f  &)  tan|  (A  +  B)  -  sin  (a  -  b)  cot£  (A  -  B). 

26.    sin(A  — B)  = 


1  -f-  cos  a  cos  6 

cos  b  —  cos  a 
1  —  cos  a  cos  6 


07           /  A    ,  T  \  sin  a  sin  b 

27.    cos(A  +  T>)  = — 

1  +  cos  a  cos  b 


28.    cos(A-B)  =  — 


1  —  cos  a  cos  6 

or»          •     o  C  -o(X  <>  b     ,  oCL     •     o  b 

29.  sin^  -  =  smj  -  cos"1  — h  cos-  -  snr  -. 

2  2222 

30.  sin  (c  -  b)  =  sin  (c  +  b)  tan2  -. 

a 

31.  sin  (a  —  6)  =  sin  a  tan  — •  —  sin  6  tan  — 

2  2 

T> 

32.  sin  (c  —  a)  =  cos  a  sin  b  tan  — 

2 

33.  If  ABC  is  a  spherical  triangle,  right-angled  at  C, 
and  cos  A  =  cos2  a,   show  that  if  A  be  not  a  right  angle, 
b  +  c  =  $7r  or  |TT,  according   as   b   and   c   are   both   <   or 

both  >  -• 


EXAMPLES. 


291 


34.  If   «,   (3  be   the   arcs    drawn   from    the   right   angle 
respectively  perpendicular  to  and  bisecting  the  hypotenuse 
c,  show  that 

sin2-  (1  -f  sin2«)  =  sin2/?. 

2t 

35.  In  a  triangle,  if  C  be  a  right  angle  and  D  the  middle 
point  of  AB,  show  that 

4  cos2-  sin2  CD  =  sin2  a  +  sin2  b. 

2i 

In  a  right  triangle,  if  p  be  the  length  of  the  arc  drawn 
from  the  right  angle  C  perpendicular  to  the  hypotenuse 
AB,  prove : 

36.  cot2p  =  cot2a  +  cot26. 

37.  cos^p  =  cos2  A  +  cos2  B. 

38.  tan2  a  =  =F  tan  a'  tan  c. 

39.  tan2  b  =  ±  tan  6'  tan  c. 

40.  tan2  a :  tan2  b  =  tan  a' :  tan  b'. 

41.  sin2/)  =  sin  a' sin  b'. 

42.  sinp  sin  c  =  sin  a  sin  b. 

43.  tan  a  tan  b  =  tan  c  sin  p. 

44.  tan2  a  +  tan2  b  =  tan2 c  cos2 p. 

45.  cot  A  :  cot  B  =  sin  a' :  sin  b'. 

In  the  oblique  triangle  ABC,  prove  the  following : 

46.  If  the  difference  between  any  two  angles  of  a  tri- 
angle is  90°,  the  remaining  angle  is  less  than  90°. 

47.  If  a  triangle  is  equilateral  or  isosceles,  its  polar  tri- 
angle is  equilateral  or  isosceles. 

48.  If  the  sides  of  a  triangle  are  each  -,  find  the  sides 
of  the  polar  triangle. 


292  SPHERICAL    TRIGONOMETRY. 

49.  If  in  a  triangle  the  side  a  =  90°,  show  that 

cos  A  +  cos  B  cos  C  =  0. 

50.  If  0  and  0'  are  the  angles  which  the  internal  and  ex- 
ternal bisectors  of  the  vertical  angle  of  a  triangle  make  with 
the  base,  show  that 

-      cos  A  ~  cos  B                         cos  A  4-  cos  B 
cos  0  = -^ ,  and  cos  0'  = — 

2  cos-  2sin- 

2  2 

51.  Given  the  base  c  and  cos      =  —  cos  C  :  find  the  locus 
of  the  vertex. 

52.  Prove  4N2  =  1  -  cos2  A  -  cos2B  -  cos2C 

—  2  cos  A  cos  B  cos  C. 

53.  If  p,  q,  r  be  the  perpendiculars  from  the  vertices  on 
the  opposite  sides,  show  that 

(1)  sin  a  ship  =  sin  b  sin  q  =  sin  c  sin  r  =  2n. 

(2)  sinAsinp  =  sin  B  sin  q  =  sinCsinr=  2N. 

54.  Prove  8?i3  =  sin2a  sin26  sin2c  sinA  sinB  sinC. 


55    Prove   s*n2^  +  sin2B  -f  sin2C  _  1  +  cos  A  cos  B  cos  G 
sin2  a  -f-  sin2  b  -\-  sin2  c         1  —  cos  a  cos  b  cos  c 

56.  If  I  be  the  length  of  the  arc  joining  the  middle  point 
of  the  base  to  the  vertex,  find  an  expression  for  its  length 

in  terms  of  the  sides.  cos  a  +  cos  b 

Ans.  cos  I  = 

2cos- 

2 

57.  If  CD,  CD'  are  the  internal  and  external  bisectors  of 
the  angle  C  of  a  triangle,  prove  that 

A.  /-.T^      cot  a  4-  cot  b        ,                     cot  a  ~  cot  b 
cot  CD  = ^-— — ,  and  cot  CD'  = 

2  cos- 


EXAMPLES.  293 

58.    Show  that  the  angles  0  and  0',  made  by  the  bisectors 

of  the  angle  C  in  Ex.  55  with  the  opposite  side  c,  are  thus 
given  : 


2sin 


cot  e,  =  sin  CD'. 

2coS| 

59.    Show  that  the  arc  intercepted  on  the  base  by  the 
bisectors  in  Ex.  55  is  thus  given  : 

sin2  A  —  sin2B 
cot  DD'  = 


2  sin  A  sinB  sinC 

60.    Prove  that 

cos2  b  —  cos2  c  cos2  c  —  cos2  a 


cos  b  cot  B  —  cos  c  cot  C      cos  c  cot  C  —  cos  a  cot  A 

cos2  a  —  cos2  b 


cos  a  cot  A  —  cos  b  cos  B 

61.  If  s  and  s'  are  the  segments  of  the  base  made  by  the 
perpendicular  from  the  vertex,  and  m  and  m'  those  made 
by  the  bisector  of  the  vertical  angle,  show  that 

s  —  s'  ,      in  —  >n'  «a  —  b 

tan— ^-  tan  — - —  =  tan'-^— 

62.  Prove 

sin  b  sin  c  4-  cos  b  cos  c  cos  A  =  sin  B  sin  C  —  cos  B  cos  C  cos  a. 

63.  Show  that  the  arc  I  joining  the  middle  points  of  the 
two  sides  a  and  b  of  a  triangle  is  thus  given : 

1  4-  cos  a  4-  cos  b  4-  cos  c 

cos  I  =  — • 1-! 

*        a        b 
4  cos  -  cos  - 

9  9 


294  SPHERICAL    TRIGONOMETRY. 

64.  If  the  side  c  of  a  triangle  be  90°,  and  8  the  arc  drawn 
at  right  angles  to  it  from  the  opposite  vertex,  show  that 

cot2  8  =  cot2  A  +  cot2  B. 

65.  Prove  that   the  angle  <£  between  the  perpendicular 
from  the  vertex  on  the  base  and  the  bisector  of  the  vertical 
angle  is  thus  given  : 


66.   In  an  isosceles  triangle,  if  each  of  the  base  angles  be 
double  the  vertical  angle,  prove  that 


cos  a  cos  ^  =  cos  f  c  +  -  )• 


a 

2          A 

67.  If  a  side  c  of  a  triangle  be  90°,  show  that 

(1)  cot  a  cot  b  +  cos  C  =  0. 

(2)  cos  S  cos  (S  -  C)  +  cos  (S  -  A)  cos  (S  -  B)  =  0. 

68.  In  any  triangle  prove 

cos  a  —  cos  b      sin  (A  —  B)  _  Q 
1  —  cos  c  sin  C 


69.  t 

=  tan  %  (a  +  b)  :  tan  |(a  —  b). 

70.  tan  J  (  A  +  a)  :  tan  £  (  A  -  a) 

=  tan  J  (B  +  b)  :  tan  £  (B  -  b). 

71.  If  the  bisector  of  the  exterior  angle,  formed  by  pro- 
ducing BA  through  A,  meet  the  base  BC  in  Df,  and  if  BD 
=  c",  CD'  =  6",  prove 

sin  b  :  sin  c  =  sin  I"  :  sin  c". 

72.  If  D  be  any  point  in  the  side  BC  of  a  triangle,  prove 

sinBD  _  sin  BAD    sin  C 
sin  CD  ~~  sin  CAD    sinB 


EXAMPLES.  295 

73.  If  A  =  a,  show  that  B  and  b  are  either  equal  or 
supplemental,  as  also  C  and  c. 

74.  If  A  =  B  -f  C,  and  D  be  the  middle  point  of  a,  show 
that  a  =  2  AD. 

75*   When   does   the   polar   triangle   coincide   with    the 
primitive  triangle  ? 

76.  If  D  be  the  middle  point  of  c,  show  that 

cos  a  +  cos  b  =  2  cos  -  cos  CD. 

77.  In  an  equilateral  triangle  show  that 

(1)  2cos-sin-  =  l. 


(2)   tan2  ~ -{- 2  cos  A  =  1. 

£l 


78.  If  b  +  c  =  TT,  show  that  sin  2  B  +  sin  2  C  =  0. 

79.  Show  that 

sin  b  sin  c  -f-  cos  b  cos  c  cos  A  =  sin  B  sin  C  —  cos  B  cos  C  cos  a. 

80.  If  D  be  any  point  in  the  side  BC  of  a  triangle,  show 
that 

cos  AD  sin  a  =  cos  c  sin  DC  +  cos  b  sin  BD. 

81.  Prove  cos2C  =  cos2^(a+6)sin2<p  +  cos2|(a-6)cos2^. 

82.  " 


83.  "      sin  s  sin  (s  —  a)  sin  (s  —  b)  sin  (s  —  c) 

=  J  (1  —  cos2  a  —  cos2  b  —  cos2  c  -f  2  cos  a  cos  b  cos  c). 

84.  If  AD  be  the  bisector  of  the  angle  A,  prove  that 

(1)  cos  B  -f-  cos  C  =  2  sin  —  sin  ADB  cos  AD. 

(2)  cos  C  -  cos  B  =  2  cos  ^  cos  ADB. 


296  SPHERICAL    TRIGONOMETRY. 

85.  Prove  cos  a  sin  b  =  sin  a  cos  b  cos  C  +  cos  A  sin  c. 

86.  "       sin  C  cos  a  =  cos  A  sin  B  -f-  sin  A  cos  B  cos  C. 

87.  In  a  triangle  if  A  =  -,   B  =  -,    C  =  -,   show   that 

5  3 


•    /&       A\      1  H- cos  a  — cos  6  — cose 
88.   Prove  sin  (S  —  A)  =  — • 

4  cos  -  sin  -  sin  - 


89.  If  8  be  the  length  of  the  arc  from  the  vertex  of  an 
isosceles  triangle,  dividing  the  base  into  segments  a  and  /8, 
prove  that 

tan  -  tan  "  =  tan  a  "*~    tan  ^^ — 


90.  If  6  =  c,  show  that 

sin  -                          cos  — 
2                                  2 
sin  6  = j-,  and  sin  B  = • 

sin-  cos- 

2  2 

91.  If  AB,  AC  be  produced  to  B',  C',  so  that  BB',  CC' 
shall   be   the   semi-supplements   of   AB?  AC  respectively, 
prove  that  the  arc  B'C'  will  subtend  an  angle  at  the  centre 
of  the  sphere  equal  to  the  angle  between  the  chords  of 
AB,  AC. 


PRELIMINARY  OBSERVATIONS.  297 


CHAPTER   XI. 
SOLUTION  OF  SPHERICAL  TRIANGLES, 

199.  Preliminary  Observations.  —  In  every  spherical  tri- 
angle there  are  six  elements,  the  three  sides  and  the  three 
angles,  besides  the  radius  of  the  sphere,  which  is  supposed 
constant.      The  solution  of  spherical  triangles  is  the  process 
by  which,  when  the  values  of  any  three  elements  are  given, 
we  calculate  the  values  of  the  remaining  three   (Art.  184, 
Note). 

In  making  the  calculations,  attention  must  be  paid  to  the 
algebraic  signs  of  the  functions.  When  angles  greater  than 
90°  occur  in  calculation,  we  replace  them  by  their  supple- 
ments ;  and  if  the  functions  of  such  angles  be  either  cosine, 
tangent,  cotangent,  or  secant,  we  take  account  of  the  change 
of  sign. 

It  is  necessary  to  avoid  the  calculation  of  very  small 
angles  by  their  cosines,  or  of  angles  near  90°  by  their  sines, 
for  their  tabular  differences  vary  too  slowly  (Art.  81).  It 
is  better  to  determine  such  angles,  for  example,  by  means 
of  their  tangents. 

We  shall  begin  with  the  right  triangle ;  here  two  ele- 
ments, in  addition  to  the  right  angle,  will  be  supposed 
known. 

SOLUTION   OF  RIGHT  SPHERICAL  TRIANGLES. 

200.  The  Solution  of  Right  Spherical  Triangles  presents 
Six  Cases,  which   may  be  solved  by  the  formulae  of  Art. 
185.     If  the  formula  required  for  any  case  be  not  remem- 
bered, it  is  always  easy  to  find  it  by  Napier's  Rules   (Art. 


298  SPHERICAL    TRIGONOMETRY. 

180).     In  applying  these  rules,  we  must  choose  the  middle 
part  as  follows : 

When  the  three  parts  considered  are  all  adjacent,  the 
one  between  is,  of  course,  the  middle  part.  When  only 
two  are  adjacent,  the  other  one  is  the  middle  part. 

Let  ABC  be  a  spherical  triangle, 
right-angled  at  C,  and  let  a,  6,  c 
denote  the  sides  opposite  the  angles 
A,  B,  C,  respectively. 

We  shall  assume  that  the  parts 
are  all  positive  and  less  than  180° 
(Art.  182). 

201.  Case  I.  —  Given  the  hypotenuse  c  and  an  angle  A ;  to 
find  a,  b,  B. 

By  (3)>  (5),  and  (8)  of  Art.  185,  or  by  Napier's  Kules, 
we  have 

sin  a  =  sin  c  sin  A, 

tan  b  =  tan  c  cos  A, 
cotB  =  cos  c  tan  A. 

Since  a  is  found  by  its  sine,  it  would  be  ambiguous,  but 
the  ambiguity  is  removed  because  a  and  A  are  of  the  same 
species  [Art.  187,  (!)]•  B  and  b  are  determined  imme- 
diately without  ambiguity. 

If  a  be  very  near  90°,  we  commence  by  calculating  the 
values  of  b  and  B,  and  then  determine  a  by  either  of  the 
formulae 

tan  a  =  sin  b  tan  A,  tan  a  =  tan  c  cos  B. 

Check.  —  As  a  final  step,  in  order  to  guard  against  numer- 
ical errors,  it  is  often  expedient  to  check  the  logarithmic 
work,  which  may  be  done  in  every  case  without  the  neces- 
sity of  new  logarithms.  To  check  the  work,  we  make  up 
a  formula  between  the  three  required  parts,  and  see  whether 


CASE  ii.  299 

it  is  satisfied  by  the  results.     In  the  present  case,  when  the 
three  parts  a,  b,  B  have  been  found,  the  check  formula  is 

sin  a  =  tan  b  cot  B   .     .     .     .     [(6)  of  Art.  185] 
Ex.  1.    Given  c  =  81°  29'  32",  A  =  32°  18'  17"  ;  find  a,  6,  B. 
Solution. 


log  sine    =9.9951945 
log  sin  A  =  9.7278843 

log  sin  a   =  9".  7230788 
.-.  a  =31°  54' 25". 

log  cose    =9.1700960 
log  tan  A  =9.8009157 

log  cot  B  =  8.9710117 
.-.  B  =  84°  39'  21". 33. 


log  tan  c    =  10.8250982 
log  cos  A  =    9.9269687 

log  tan  b   =10.7520669 
.-.  6   =  79°  51' 48".65. 

Check. 

log  tan  b   =10.7520669 
log  cot  B  =    8.9710117 

log  sin  a  =    9.7230786' 


Ex.  2.    Given  c  =  110°  46'  20",  A  =  80°  10'  30" ;  find  a,  6,  B. 
Ans.  a  =  67°  5'  52". 7,  b  =  155°  46'  42". 7,  B  =  153°  58'  24".5. 

202.  Case  II.  —  Given  the  hypotenuse  c  and  a  side  a  ;  to 
find  6,  A,  B. 

By  (1),  (3),  (4)  of  Art.  185,  or  by  Napier's  Rules,  we 
have 

,      cos  c     •     A       sin  a         -p,      tan  a 

cos  b  = ,  sm  A  =  —  — ,  cos  B  =  —  — 

cos  a  sin  c  tan  c 

The  check  formula  involves  6,  A,  B ;  therefore,  from  (9) 
of  Art.  185  we  have 

cos  B  =  sin  A  cos  b. 

In  this  case  there  is  an  apparent  ambiguity  in  the  value 
of  A,  but  this  is  removed  by  considering  that  A  and  a  are 
always  of  the  same  species  (Art.  187). 


300 


SPHERICAL    TRIGONOMETRY. 


Ex.  1.    Given  c  =  140°,  a  =  20° ;  find  6,  A,  B. 
Solution. 


log  cose      =  9.8842540- 
colog  cos  a  =  0.0270142 

log  cos  b     =  9.9112682- 
.-.  b     =  144°  36'  28".4. 

log  tan  a     =9.5610659 
colog  tan  c  =  0.0761865- 

logcosB    =  9.6372524- 
.-.  B    =  115°  42'  23".8. 


log  sin  a     =  9.5340517 
colog  sine  =  0.1919325 

log  sin  A    =9.7259842 
.-.  A    =32°8'48".l. 

Check. 

log  sin  A    =9.7259842 
log  cos  b     =  9.9112682 

log  cos  B    =9.6372524 


Ex.  2.    Given  c  =  72°  30',  a  =  45°  15' ;  find  b,  A,  B. 

Ans.  b  =  64°  42'  52",  A  =  48°  7' 44".5,  B  =  71°  27' 15". 

203.   Case  III. — Given  a  side  a  and  the  adjacent  angle  B ; 
to  find  A,  6,  c. 

By  (10),  (6),  (4)  of  Art.  185,  we  have 
cos  A  =  cos  a  sin  B,  tan  b  =  sin  a  tan  B,  cot  c  =  cot  a  cos  B. 

Check  formula,          cos  A  =  tan  b  cot  c. 
In  this  case  there  is  evidently  no  ambiguity. 
Ex.  1.    Given  a =31° 20' 45",  B  =  55°30'30";  find  A,  6,  c. 
Solution. 


log  cos  a  =9.9314797 
log  sin  B  =9.9160371 

log  cos  A  =  9.8475168 
.-.  A  =  45°  15' 30". 6. 

log  cot  a  =0.2153073 
log  cos  B  =  9.7530361 

log  cote   =9.9683434 
.-.   c   =47°  5' 11". 


log  sin  a  =9.7161724 
log  tan  B  =0.1630010 

log  tan  6  =9.8791734 
.-.  b  =37°  7' 50". 

Check. 

log  tan  b  =  9.8791734 
log  cote   =9.9683434 

log  cos  A  =  9.8475168 


CASE  IV.  301 

Ex.  2.    Given  0=112°  0'0",B  =  152°  23'  1".3;  find  A,  6,  c. 
Ans.  A  =  100°,  b  =  154°  7'  26".5,  c  =  70°  18'  10".2. 

204.   Case  IV.  —  Given  a  side  a  and  the  opposite  angle  A  ; 
to  find  b,  c,  B. 

BJ  (7)>  (3)>  (10)  of  Art.  185,  we  havev 

sin  6  =  tan  a  cot  A,  sinc  =  ^A  sin  B  = 


sin  A  cos  a 

(7/iecfc  formula)  sin  6  =  sin  c  sin  B. 

In  this  case  there  is  an  ambiguity,  as  the  parts  are  deter- 
mined by  their  sines,  and  two  values  for  each  are  in  general 
admissible.  But  for  each  value  of  b  there  will,  in  general, 
be  only  one  value  for  c,  since  c  and  b  are  connected  by  the 
relation  cos  c  =  cos  a  cos  b  (Art.  185);  and  at  the  same  time 
only  one  admissible  value  for  B,  since  cos  c  =  cot  A  cot  B. 
Hence  there  will  be,  in  general,  only  two  triangles  having 
the  given  parts,  except  when  the  side  a  is  a  quadrant  and 
the  angle  A  is  also  90°,  in  which  case  the  solution  becomes 
indeterminate. 

It  is  also  easily  seen  from  a  figure  that  the  ambiguity 
must  occur  in  general  (Art.  188). 

When  a—  A,  the  formulae,  and  also  the  figure,  show  that 
b,  c,  and  B  are  each  90°. 

Ex.  1.   Given  a  =  46°  45',  A  =  59°  12'  ;  find  b,  c,  B. 
Solution. 


log  tan  a  =  0.0265461 
log  cot  A  =9.7753334 

log  sin  b  =9.8018795 
.-.  b  =39°19'23".5, 
or  140°40'36".5. 


log  sin  a   =9.8623526 
log  sin  A  =9.9339729 

log  sine    =9.9283797 

.-.  c    =57°  59' 29", 

or  122°   0'31". 


302  SPHERICAL    TRIGONOMETRY. 


log  cos  A  =9.7093063 
log  cos  a  =9.8358066 

log  sin  B  =9.8734997 
.-.  B  =48°  21'  28", 


Check. 

log  sine    =9.9283797 
log  sin  B  =  9.S734997 

log  sin  b   =9.8018794 


or  131°  38'  32". 
Ex.  2.   Given  a=  1.12°,  A  =  100°:  find  &,  c,  B. 
^t/is.  &  =  154°   7'26".5,    c  =  70°18'10".2,    B  =  152°23'    1".3, 
or     25°52'33".5,   or   109°41r49".8,    or     27°36'58".7. 

205.   Case  V.  —  Given  the  two  sides  a  and  b  ;  to  find  A, 
B,c. 

By  (7)>  (6)>  (!)  of  Art  185>  we  have 

cot  A  =  cot  a  sin  b,  cot  B  =  cot  b  sin  a,  cos  c  =  cos  a  cos  b. 
Check  formula,          cos  c  =  cot  A  cot  B. 
In  this  case  there  is  no  ambiguity. 

Ex.  1.    Given  a  =  54°  16',  &  =  33°12';  find  A,  B,  c. 

Ans.  A  =  68°  29'  53",  B  =  38°  52'  26",  c  =  60°44'46". 
.Ex.  2.    Given  a  =  56°  34',  b  =  27°  18';  find  A,  B,  c. 

.  A  =  73°  9'  13",  B  =  31°  44'  9",  c  =  60°  41'  9". 


206.   Case  VI.  —  Given  the  two  angles  A  and  B  ;  to  find  a, 
b,  and  c. 

By  (10),  (9),  (8) 

cos  a  =  ^A  cos  b  =  ^-5,  cos  c  =  cot  A  cot  B. 
sin  B  sin  A 

Check  formula,        cos  c  =  cos  a  cos  b. 
There  is  no  ambiguity  in  this  case. 

Ex.  1.    Given  A  =  74°  15',  B  =  32°  10'  ;  find  a,  6,  c. 

Ans.  a  =  59°  20'  44",  b  =  28°  24'  54",  c  =  63°  21'  24".5. 

Ex.  2.    Given  A  =  91°  11',  B  =  111°  11'  ;  find  a,  b,  c. 

Ans.  a  =  91°  16'  8",  b  =  111°  11'  16",  c  =  89°  32'  29". 


EXAMPLES. 


303 


207.  Quadrantal  and  Isosceles  Triangles.  —  Since  the 
polar  triangle  of  a  quadrantal  triangle  is  a  right  triangle 
(Art.  184),  we  have  only  to  solve  the  polar  triangle  by  the 
formulae  of  Art.  185,  and  take  the  supplements  of  the  parts 
thus  found  for  the  required  parts  of  the  given  triangle ;  or 
we  can  solve  the  quadrantal  triangle  immediately  by  the 
formulae  of  Art.  189.* 

A  biquadrantal  triangle  is  indeterminate  unless  either 
the  base  or  the  vertical  angle  be  given. 

An  isosceles  triangle  is  easily  solved  by  dividing  it  into 
two  equal  right  triangles  by  drawing  an  arc  from  the  vertex 
to  the  middle  of  the  base. 

The  solution  of  triangles  in  which  a  +  b  =  TT,  or  A  +  B  =  TT, 
can  be  made  to  depend  on  the  solution  of  right  triangles. 
Thus  (see  the  second  figure  of  Art.  191)  the  triangle  B'AC 
has  the  two  equal  sides,  a'  and  b,  given,  or  the  two  equal 
angles,  A  and  B',  given,  according  as  a  -f-  b  =  TT  or  A  +  B  =  IT 
in  the  triangle  ABC. 

EXAMPLES. 
Solve  the  following  right  triangles  : 


1. 


4. 


Given  c=32°34f, 
find  a=22°15'43", 

Given  c= 69°  25' 11", 
find  a  =  50°   0'  0", 

Given  c=55°   9' 32", 
find   6=51°53', 

Given  c=127°12', 
find   6=39°   6' 25", 

Given  a  =  118°  54', 
find  A =95°  65'   2", 


a=44°44'; 
6  =  24°  24'  19", 

A  =  54°  54'  42"; 
b  =  56°  50'  49", 

a=22°15'   7"; 
A=27°28'37'.5, 


A=128°5'54", 


=  10°  49'  17", 


B  =  50°8'21". 


=  63°25'4". 


=  52°21'49". 


c= 118°  20'  20". 


*  Quadrantal  triangles  are  generally  avoided  in  practice,  but  when  unavoidable, 
they  are  readily  solved  by  either  of  these  methods. 


304 
6. 

7. 


SPHERICAL    TRIGONOMETRY. 


1.0. 


11. 


12. 


Givena=29°46'   8", 
findA=54°   1'16", 

Given  a= 77°  21' 50", 
find   6=28°14'31".l, 
or     6'=151°45'28".9, 

Given  a =68°, 
find   6=25°52'33".5, 
or      6'  =  154°7'26".5, 

Given  a  =  144°  27' 3", 
find  A  =  126°  40' 24", 

Given  a =36°  27', 
find  A  =  46059'43".3, 

Given  A  =  63°  15' 12", 
find  a  =  49°  59'  56", 

Given  A  =  67°  54' 47",  • 
find  a= 67° 33' 27", 


B  =  137°24'21"; 

6  =  155°  27'54"? 

c=  142°  9'  13". 

A=  83°  56'  40"; 

c=78°53'20", 

B=28°49'57".4, 

c'  =  101°6'40", 

B'=151°10'2".G. 

c=70°18'10".2, 

B=27°36'58".7, 

c'=109°41'49".8, 

B'  =  152°23'l".3. 

B  =  47°  13'  43", 

c=  133°  32'  26". 

6  =  43°32'31"; 

B  =  57°59'19".2, 

c=54°20'. 

B  =  135°  33'  39"; 

6=143°  5'  12", 

c=  120°  55'  34". 

B  =  99°  57'  35"; 

6  =  100°  45', 

c=94°5'. 

13.  Solve  the  quadrantal  triangle  in  which 

c  =  90°,  A  =  42°l',  B  =  121°20'. 
Ans.  C  =  67°16'22",  6  =  112°10'20", 

14.  Solve  the  quadrantal  triangle  in  which 

a  =  174°12'49".l,  b  =  94° 8' 20",  c  =  ( 
Ans.  A  =  175°57'10",  B  =  135°42'55", 


=  46°31'30". 


=  135°34'8". 


SOLUTION   OF  OBLIQUE  SPHERICAL  TRIANGLES. 

208.  The  Solution  of  Oblique  Spherical  Triangles  presents 
Six  Cases  ;  as  follows  : 

I.    Given  two  sides  and  the  included  angle,  a,  b,  C. 
II.    Given  two  angles  and  the  included  side,  A,  B,  c. 
III.    Given  two  sides  and  an   angle   opposite  one  of  them, 
a,  b,  A. 


CASE  I.  305 

IV.  Given  two  angles  and  a  side  opposite  one  of  them, 
A,  B,  a. 

V.    Given  the  three  sides,  a,  b,  c. 
VI.    Given  the  three  angles,  A,  B,  C. 

These  six  cases  are  immediately  resolved  into  three  pairs 
of  cases  by  the  aid  of  the  polar  triangle  (Art.  184)  . 

For  when  two  sides  and  the  included  angle  are  given, 
and  the  remaining  parts  are  required,  the  application  of 
the  data  to  the  polar  triangle  transforms  the  problem  into 
the  supplemental  problem:  given  two  angles  and  the 
included  side,  to  find  the  remaining  parts. 

Similarly,  cases  III.  and  IV.  are  supplemental,  also  V. 
and  VI. 

The  parts  are  all  positive  and  less  than  180°  (Art.  182). 
The  attention  of  the  student  is  called  to  Art.  199. 

209.  Case  I.  —  Given  two  sides,  a,  b,  and  the  included 
angle  C  ;  to  find  A,  B,  c. 

By  Napier's  Analogies,  (7)  and  (8)  of  Art.  197, 
tan  *  (A  +  B)  =  cos      a-  6         C 


cos  i  (a  4-  &)        2 


. 

sin  \  (a  4-  6)        2 

These  determine  ^(A  +  B)  and  -J-(A  —  B),  and  there- 
fore A  and  B  ;  then  c  can  be  found  by  Art.  190,  or  by  one 
of  Gauss's  equations  (Art.  198)  .  Since  c  is  found  from  its 
sine  in  Art.  190,  it  may  be  uncertain  which  of  two  values  is 
to  be  given  to  it  :  if  we  determine  c  from  one  of  Gauss's 
equations,  it  is  free  from  ambiguity.  We  may  therefore 
find  c  from  (3)  of  Art.  198.  Thus 

cos  £  (A  +  B)  cos-  =  cos  £  (a  4-  &)  sin^- 

2 


306 


SPHERICAL    TRIGONOMETRY. 


Check,  tan  £  (a  +  6)  cos  £  (A  +  B)  =  cos  $•  (A  -  B)  tan  |. 
There  is  no  ambiguity  in  this  case. 

Ex.  1.   Given   a  =  43°  18',    b  =  19°  24',    C  =  74°  22' ;    find 
A,  B,  c. 

Solution. 

£  (a  +  6)  =  31°  21',  £  (a  -  6)  =  11°  57',  £  C  =  37°  11'. 


log  cos  £  (a  — 6)  =  9.9904848 
logseci(a  +  6)  =  0.0685395 

log  cot  5  =10.1199969 

2      


logtan4(A+B)  =10.1790212 


0".5 


.-.  A  =  84°  10'  17".5 

B  =  28°  48'  16".5. 
c  =  41°35'48".5. 


log  sin  4-  (a  -6)=  9.3160921 
logcosec£(a+&)  =  0.2837757 

log  cot  5=io.  1199969 
logtan|(A-B)  =  9.7198647 
...  i(A-B)=27°41'0".5. 

log  cos  |  (a  +  6)  =  9.9314605 
logseci(A  +  B)  =  0.2579737 

log  sin  -  =  9.7813010 


log  cos  -=  9.9707352 


.-.  -  =  20°  47'  54".25. 

2 


Otherwise  thus:  Let  fall  the 
perpendicular  BD,  dividing  the 
triangle  ABC  into  two  right  tri- 
angles, BDA,  BDC.  Denote  AD 
by  ra,  the  angle  ABD  by  <£,  and  A 
BD  by  p.  Then  by  Napier's  Rules, 
we  have 

cos  C  =  tan  (b  —  m)  cot  a ; 

sin  (b  —  m)  =  cot  C  tanp ;  sin  m  =  cot  A 

.-.  tan(6  —  m)  =  tan  a  cos  C (1) 

and        tan  A  sin  m  =  tan  C  sin  (b  —  m) (2) 


CASE  II.  307 

From  (1)  m  is  determined,  and  from  (2)  A  is  determined. 
In  a  similar  manner  B  may  be  found.  * 

Also,  from  the  same  triangles,  we  have  by  Napier's  Kules 

cos  a  *=  cos  (b  —  m)  cosp  ;  cos  c  =  cos  m  cos  p. 
.'.  cos  c  cos  (6  —  m)  =  cos  m  cos  a, 
from  which  c  is  found. 

NOTE.  —  This  method  has  the  advantage  that,  in  using  it,  nothing  need  be  remem- 
bered except  Napier's  Rules. 

If  only  the  side  c  is  wanted,  it  may  be  found  from  (4)  of  Art.  191,  without  pre- 
viously determining  A  and  B.  This  formula  may  be  adapted  to  logarithms  by  the 
use  of  a  subsidiary  angle  (Art.  90). 

Ex.2.  Given  6  =  120°  30'  30",  c=70°20'20",  A=50°iOr10"; 
find  B,  C,  a. 

Ans.  B  =  135°  5'  28".8,  C  =  50°  30'  8%  a  =  69°  34'  56". 

210.  Case  II.  —  Given  two  angles,  A,  B,  and  the  included 
side  c  ;  to  find  a,  6,  C. 

By  Napier's  Analogies  (9)  and  (10)  of  Art.  197, 


cosi(A4-B)        2 


sm£(A  +  B)        2 
from  which  a  and  b  are  found. 

The  remaining  part  C  may  be  found  by  (2)  of  Art.  198. 
sin  %(a  —  b)  cos  ~  =  sin  |  (A  —  B)  sin  -• 

Check,    cos  J  (a  -  b)  cot  5  =  cos  £  (a  +  b)  tan  £  (  A  +  B)  . 
There  is  no  ambiguity  in  this  case. 


308 


SPHERICAL   TRIGONOMETRY. 


Ex.  1.  Given  A =68°  40',  B  =  56°20f,  c=84°30';  find  a,  b,  C. 

Solution. 
£  (A  +  B)  =  62°  30',  i  (A  -  B)  =  6°  10',  -  =  42°  15'. 


logcosi(A-B)  =  9.9974797 
logseci(A  +  B)  =  0.3355944 

log  tan -=   9.9582465 


log  tan  ^(a  +  b)  =10.2913206 

.'-.  i(a  +  6)  =  62°55'    9" 

|(a-&)=    6°  16' 39" 

a  =  69° 11' 48" 

b  =  56°  38'  30". 
0  =  97°  19'   3".5. 


logsini(A-B)   =9.0310890 
logcoseci(A+B)  =0.0520711 

log  tan -=9.9582465 

2     

log  tan  i(a  -  6)  =9.0414066 
...  -i-  (a  -b)  =6°  16'  39". 

log  sin  £(A-B)  =9.0310890 
log  cosec  i(a  -  b)  =0.9612050 

log  sin -=9.8276063 
log  cos -=9.8199003 
.  5  _  48°  39'  31f ". 

a 


Otherwise  thus :  Let  fall  the  perpendicular  BD  (see  last 
figure).  Denote,  as  before,  AD  by  972,  the  angle  ABD  by  <£, 
and  BD  by  p.  Then  by  Napier's  Eules,  we  have 

cos  c  =  cot  <£  cot  A ; 
cos  <£  =  cote  tan  jo;  cos(B  —  <£)  =  cot  a  tan^>. 

.-.  cot  <£  =  tan  A  cos  c (1) 

tan  a  cos  (B  —  <£)=cos<£  tanc (2) 

From  (1)  <f>  is  determined,  and  from  (2)  a  is  found. 
Similarly  b  may  be  found. 

Also,  from  the  same  triangles,  we  have 

cos  C  sin  $  =  cos  A  sin(B  —  <£), 
from  which  C  is  found. 


CASE  111.  309 

Ex.  2.    Given 

A  =  135°  5'  28".C,  C  =  50°  30'  8".6,  b  =  69°  34'  56".2  ; 
find  a,  c,  B. 

Ans.  a  =  120°  30'  30",  c  =  70°  20'  20",  B  =  50°  10'  10". 

211.   Case  III.  —  Given  two  sides,  a,  b,  and  the  angle  A 
opposite  one  of  them  ;  to  find  B,  C,  c. 

The  angle  B  is  found  from  the  formula, 


sin  B=-  sin  A     .....     (Art.  190)  (1) 

sin  a 

Then  C  and  c  are  found  from  Napier's  Analogies, 


^,,     ,  sin  A  _  sinB  _  sinC 

sin  a       sin  b       sin  c 

Since  B  is  found  from  its  sine  in  (1),  it  will  have  two 
values,  if  sin  A  sin  b  <  sin  a,  and  the  triangle,  in  general, 
will  admit  of  two  solutions. 

When  sin  A  sin  b  >  sin  a,  there  will  be  no  solution,  for 
then  sin  B  >  1. 

In  order  that  either  of  these  values  found  for  B  may  be 
admissible,  it  is  necessary  and  sufficient  that,  when  sub- 
stituted in  (2)  and  (3),  they  give  positive  values  for 

tan  —  and  tan  -,  or  which  is  the  same  thing,  that  A  —  B  and 
a  —  b  have  the  same  sign.     Hence  we  have  the  following 

Rule.  —  If  both  values  of  B  obtained  from  (1)  be  such  as 
that  A  —  B  and  a  —  b  have  like  signs,  there  are  two  complete 
solutions.  If  only  one  of  the  values  of  B  satisfies  this  condi- 
tion, there  is  only  one  triangle  that  satisfies  the  problem,  since 


310 


SPHERICAL   TRIGONOMETRY. 


in  this  case  C,  or  c>180°.  If  neither  of  the  values  of  B 
makes  A  —  B  and  a  —  b  of  the  same  signs,  the  problem  is 
impossible. 

This  case  is  known  as  the  ambiguous  case,  and  is  like  the 
analogous  ambiguity  in  Plane  Trigonometry  (Art.  116), 
though  it  is  somewhat  more  complex.  For  a  complete 
discussion  of  the  Ambiguous  Case,  the  student  is  referred 
to  Todhunter's  Spherical  Trigonometry,  pp.  53-58 ;  McCol- 
lend  and  Preston's  Spherical  Trigonometry,  pp.  137-143; 
Serret's  Trigonometry,  pp.  191-195,  etc. 

Ex.1.  Given  a=42° 45',  6  =  47°  15',  A  =  56° 30';  findB,C,c. 


Solution. 


log  sin  b  =  9.8658868 

cologsiria  =0.1682577 

log  sin  A  ='9.9211066 

log  sin  B  =9.9552511 
.-.  B  =  64°26'4", 
B'  =  115°  33'  56". 


+  B)  = 
|(A-B)  = 


45°    0'    0". 

-  2°  15'    0". 
60°  28'   2". 

-  3°  58'    2". 
86°    1'68". 

-29°31'5S". 


Since  both  values  of  B  are  such  that  A  —  B,  A  —  B',  and 
a  —  b,  are  all  negative,  there  are  two  solutions,  by  the  above 
Rule. 


(1)   When  B  =  64°  26'  4". 

logsinJ(o-6)   =  8.5939483- 
colog  sin  i  (a +  5)  =0.1505150 
log  cot  |(A  -  B)  =  1.1589413- 

log  tan -=9.9034046 

...  £  =  ,38°  40'  48". 
.-.  0  =  77°  21' 36". 


logsini(A  +  B)  =9.9395560 
cologsini(A-B)  =  1.1599832- 
log  tan  i(a  -  b)    =8.5942832- 

log  tan  -=9.6938224 


.-.  £=26°  17'  40". 
.-.  c=  53°  35'  20". 


CASE  III. 


311 


(2) 

logsini(a-6)    =8.5939483- 
cologsin£(a+6)  =0.1505150 
log  cot  i(A-B')  =  0.2467784- 

log  tan  ®  =8.9912417 

.•..5-'=  5°35'50J". 
.-.  C'=ll0H'40i". 


B')   =9.9989581 
cologsm-J(A-B')  =  0.3072223- 
log  tan|(a  -  !>)  =8.5942832 - 


log  tan -=8.1 
o 


9004636 


.-.  -  =  4°32'47i". 
.-.  c'=9°    5'34". 


?.  B  =    64°  26'    4",  C  =77°  21'  36",     c  =53°  35' 20"; 
B' =  115°  35' 56",  C'  =  ll°ll'40£",  c'=    9°    5' 

Otherwise  thus:  Let  fall  the 
perpendicular  CD ;  denote  AD 
by  m,  the  angle  ACD  by  <£,  and 
CD  by  p.  Then  we  have 

cos  A  =  tan  m  cot  b  ; 
.-.  tan  m  =  cos  A  tan  b      (1) 


cos  b  =  cot  A  cot  0.     .  •.  cot  <f>  =  cos  b  tan  A 
Again, 

cos  a  =  cos  (c  —  m)  cos  p ;  cos  b  =  cos  m  cos  p. 

.-.  cos  (c  —  m)  =  cos  a  cos  m-^-b     .     . 
Also,  cos  (C  —  <£)  =  cot  a  tan  p ;  cos  <£  =  cot  b  tanp. 
.-.  cos  (C  —  (j>)  =  cot  a  tan  6  cos  <£  .     . 

Lastly, 


(2) 


(4) 


sin  a 


The  required  parts  are  given  by  (1),  (2),  (3),  (4),  (5). 

Ex.  2.  Given  a  =  73°49'38",  6=120° 53' 35",  A=88°52'42": 
find  B,  C,  c. 

Ans.  B  =  116°  44' 48",  C  =  116°  44' 48",  c  =  120°  55' 35". 


812 


SPHERICAL   TRIGONOMETRY. 


212.  Case  IV.  —  Given  two  angles,  A,  B,  and  the  side  a 
opposite  one  of  them ;  to  find  b,  c,  C. 

This  case  reduces,  by  aid  of  the  polar  triangle,  to  the 
preceding  case,  and  gives  rise  to  the  same  ambiguities. 
Hence  the  same  remarks  made  in  Art.  211  apply  in  this 
case  also,  and  the  direct  solution  may  be  obtained  in  the 
same  way  as  in  Case  III.  Thus, 

The  side  b  is  found  from  the  formula 


sin  b  = 


sinB 

sin  A 


sin  a (1) 


Then  c  and  C  are  found  from  Napier's  Analogies. 

ta    c      cosHA  +  B)  _ 

2      cosf  (A  —  B) 


Check, 


2 
sin  A 


cos  %  (a  +  b) 
sin  B      sin  C 


(2) 
(3) 


sin  a       sin  b       sin  c 

Ex.  1.    Given  A=66°7'20",  B=52°50'20",  a=59°28'27"; 

find  6,  c,  C. 

Solution. 

By  (1)  we  find  b  =  48°  39'  16",  or  131°  20' 44". 

i(A  +  B)=59°28'50". 
i(A-B)  =  6°  38' 30". 

log  cos  |(A  +  B)  =  9.7057190 


cologcosi(A-B)=  0.0029244 
log  tan  J  (a  +  b)  =  10.1397643 


log  tan -  =   9.8484077 


-  =  35°  II1 


%(a-b)=    5°  24'  351". 


log  cos  %  (a  -  b)  =9.9980612 
cologcos^(a  +  &)  =0.2314530 


log  cot  J(A  +  B)=  9.7704854 
log  tan  -  =9.9999996 


=  45'. 


The  second  value  of  b  is  inadmissible  (see  Eule  of  Art. 
211),  and  therefore  there  is  only  one  solution. 

Ans.  b  =  48°  39'  16",  c  =  70°  23'  4H",  C  =  90°. 


CASE  V.  313 

Ex.  2.  Given  A  =  110°  10',  B  =  133°  18',  a  =  147°  5'  32" ; 
find  6,  c,  C. 

-4ns.  b  =  155°  5'  18",  c  =  33°  1'  45",  C  =  70°  20'  50". 

213.  Case  V.  —  Given  the  three  sides,  a,  &,  c ;  to  find  the 
angles. 

The  angles  may  be  computed  by  any  of  the  formulae  of 
Art.  195 ;  but  since  an  angle  near  90°  cannot  be  accurately 
determined  by  its  sine,  nor  one  near  0°  by  its  cosine  (Art. 
151),  neither  of  the  first  six  formulae  can  be  used  with  advan- 
tage in  all  cases.  The  formulae  for  the  tangents  however  are 
accurate  in  all  parts  of  the  quadrant,  and  are  therefore  to 
be  preferred  for  the  solution  of  a  triangle  in  which  all  three 
sides  or  all  three  angles  are  given. 

By  (7)  of  Art.  195  we  have 


tan  -        /sin  0  ~  6)  sin  0  -  c) 
2       \      sin  s  sin  (s  —  a) 


—          1  /sin  (s  —  a)  sin  (s  —  b)  sin  (s  —  c) 

siu(s  —  a)\  sins 

Since  the  part  under  the  radical  is  a  symmetric  function 
of  the  sides,  it  is  in  the  formulae  for  determining  all  three 
angles  A,  B,  C,  and  when  once  calculated,  it  may  be  utilized 
in  the  calculation  of  each  angle.  For  convenience  in  com- 
putation, denote  this  term  by  tan  r.  Then 


tanr=     /sin  (s  -  a)  sin  (s  -  b)  sin  (s  -  c)  . 
\  sins 

and  (7),  (8),  (9)  of  Art.  195  become 

(1) 


sin  (s  -  a) 
tanr 


(2) 


2      sin  (s  -  b) 

g         *£!L^          .......     (3) 

2       sin  (s  —  c) 


314 

Check, 


SPHERICAL   TRIGONOMETRY. 

sin  A      sin  B      sin  C 


sma 


sinB 
sin  6 


sine 


Ex.1.    Given    a  =  46°  24',    6  =  67°  14',   c  =  81°12';    ftnd 
A,  B,  C. 

Solution. 


a  =    46°  24' 

b=    67°  14' 

c  =    81°  12' 


2s  =  194°  50' 

s=  97°  25', 

§-a=  51°    1', 

*_&  =  30°  11', 

s-c  =  16°  13'. 


log  sin  (s  -  a)  =  9.8906049 

log  sin  (8  -b)  =  9.7013681 

log  sin  (s  -  c)  =  9.4460251 

colog  sin  s  =  0.0036487 

log  tan2  r  =  9.0416468 
log  tan  r=  9.5208234. 


tan  r  =  9.5208234 
sin(s-a)  =  9.8906049 


tan -=9.6302185 

2 

.-.  -=23°    6' 45". 
A  =46°  13' 30". 


tan  r  =  9.5208234 
sin(«-6)  =  9.7013681 


tan -=9.8194553 

2 

.-.  ?=  33°  25' 10". 
B=  66°  50' 20". 


tanr=  9.5208234 
sin(s-c)=  9.4460251 

tan -=10.0747983 

2 

X  ~  =49°  54' 35". 
A 

C=  99°  49' 10". 


Ex.  2.  Given  a  =  100°,  b  =  37°  18',  c  =  62°  46' ;  find  A, 
B,  C. 

Ans.  A  =  176°15'46".56,  B  =  2°17'55".08,  C  =  3°22'25".46. 

214.  Case  VI.  — Given  the  three  angles,  A,  B,  C;  to  find 
the  sides. 

As  in  Art.  213,  the  formulae  for  the  tangents  are  to  be 
preferred. 

Putting   tan  B  = 

we  have,  from  (7),  (8),  (9)  of  Art.  196, 

tan  ^  =  tan  R  cos  (S  —  A), 


CASE   VI. 


315 


tan  -  =  tan  R  cos  (S  -  B), 

2 

tan  £  =  tan  K  cos  (S  -  C), 

Zi 

by  which  the  three  sides  may  be  found, 
sin  A      sin  B      sin  C 


Check, 


sin  a 


sin  b 


sine 


Ex.1.   Given   A  =  68°  30',  B  =  74°20',  C  =  83°10';   find 
a,  fe,  c. 

Solution. 


A  =    68°  30' 
B=    74°  20' 

C  =    83°  10' 
28  =  226°   0' 

log  (-  cos  S)  =  9.5918780 

log  cos  (S  -  A)  =  9.8532421 

log  cos  (S  -  B)  =  9.8925365 

log  cos  (S  -  C)  =  9.9382576 

log  tan2  R  =  9.9078418 

log  tan  E  =  9.9539209.* 


8  =  113°   0' 

S-A  =    44°  30' 
S-B=    38°  40' 

S  -  C  =    29°  50' 


log  tan -  =  9.8071630 

a 

log  tan  |  =  9.8464574 
log  tan -  =  9.8921785 


a  =  65°  2 

&  =  70°    9'    9i 

c  =  75°55'    9". 


Check, 


sin  a 


sin  b 


sine 


sin  A      sin  B      sin  C 


Ex.2.  Given  A=59°55'10",  B  =  85°36'500,  C  =  59°55'10"; 
find  a,  b,  c. 

Ans.  a  =  129°  11' 40",  b  =  63°  15'  12",  c  =  129°  11' 40". 

*  The  necessary  additions  may  be  conveniently  performed  by  writing  log  tan  R  on 
a  slip  of  paper,  and  holding  it  successively  over  log  cos  (S  — A),  log  cos  (S  —  B),  and 
logcos(S-C). 


316 


SPHERICAL   TRIGONOMETRY. 


EXAMPLES. 

Solve  the  following  right  triangles  : 

Given  c=   84° 20', 
find    a=  35°  13'  4", 

Given  c=   67°  54', 
find   a=  39°35'51", 

3.  Given  c=  22°18'30", 

find   a=   16°  17' 41", 

4.  Given  c  =  145°, 

find   a=  13°  12' 12", 

5.  Given  c=  98°   6' 43", 

find   a=137°   6', 

6.  Given  c=  46°  40' 12", 

find   a=  26°27'23".8, 

7.  Given  c=  76°  42', 

find   b=  70°  10' 13", 

8.  Given  c=   91°  18', 

find    6=  94°  18' 53 ".8, 

9.  Given  c=  86°51', 

find   a=  86°  41 '14", 

10.  Given  c=  23°49'51", 

find   6=  19°  17', 

11.  Given  c=  97°  13'  4", 

find    b=  79°13'38".2, 

12.  Given  c=  37°40'20", 

find    6=     0°26'37".2, 


A=  35°  25'; 

6=  83°   3'  29", 

B=  85°  59'   1". 

A=  43°28'; 

b=  60°  46'  25J", 

B=  70°22'21". 

A=  47°  39'  36"  5 

b=  15°26'53", 

B=  44°33'53".4. 

A=  23°  28'; 

6=147°17'15", 

B  =  109°  34'  33". 

A  =  138°  27'  18"; 

b=  77°51', 

B=  80°55'27". 

A=  37°  46'   9"; 

b=  39°57'41".4, 

B=  62°  0'  4". 

a=  47°18'; 

A=  49°  2'24".5, 

B==  75°  9'  24  ".75. 

a=   72°  27'; 

A=  72°  29'  48", 

B=  94°  6'53".3. 

6=   18°   1'50"; 

A=  88°  58'  25", 

B=  18°  3'  32". 

a=   14°16'35"; 

A=  37°36'49".3, 

B=  54°49'23".3. 

a=  132°  14'  12"; 

A  =  131°  43'  50", 

B=  81°58'53".3. 

a=   37°  40'  12"; 

A-:  89°25'37", 

B=     0°43'33". 

EXAMPLES.                                             317 

Given  a=  82°   6', 

B=   43°  28'; 

find  A=  84°  34'  28", 

b=  43°  11'  38", 

c=  84°  14'  57". 

Given  a=  42°30'30", 

B=  53°  10'  30"; 

find  A=  53°  50'  12", 

b=  42°   3'  47", 

c=  56°  49'  8". 

Given  a=  20°20'20", 

B=  38°  10'  10"; 

find  A=  64°35'16".7, 

b=  15°16'50".4, 

c=  25°  14'  38  ".2. 

Given  a=  92°  47'  32"; 

B=   50°   2'   1"; 

find  A=  92°  8'  23", 

b=  50°, 

c=  91°47'40". 

Given  b=  54°30', 

A=  35°  30'; 

find  B=  70°17'35", 

a=  30°   8'39".2, 

c=  59°  51'  20". 

Given  b=  155°  46'  42".  7, 

A-  80°10'30"; 

find  B  =  153°68'24".5, 

a=   67°   6'52".6, 

c=110°46'20". 

Given  a=   35°  44', 

A=  37°  28'; 

find    6=  69°  50'  24", 

c=  73°  45'  15", 

B=   77°  54', 

or      &'=110°   9'  36", 

c'=  106°  14'  45", 

B  =  102°  6'. 

Given  a=129°33', 

A  =  104°  59'; 

find    b=  18°54'38", 

c=127°  2'  27", 

B=  23°57'19", 

or      6'=161°  5'  22", 

c'=  52°  57'  33", 

B'  =  156°  2'  41". 

Given  a=  21°  39', 

A=  42°  10'  10"; 

find    6=  25°59'27".8, 

c=  33°20'13".4? 

B=  52°  23'  2".8, 

or      6'=154°   0'32".2, 

c'=146°39'46".6, 

B'=127°36'57".2. 

Given  a=  42°  18'  45", 

A=   46°15'25"; 

find    6=  60°  36'  10", 

c=   68°  42'  59", 

B=  69°13'47", 

or      &'=119°23'50", 

c'=lll°17'   1", 

B'  =  110°46'13". 

Given  &  =  160°, 

B  =  150°; 

find   a=  39°   4'50".7, 

c=136°50f23".3, 

A=  67°  9'42".7, 

or     a'=140°55f  9".3, 

c'=  43°  9'36".7, 

A'=112°50'17".3. 

318                           SPHERICAL    TRIGONOMETRY. 

24. 

Given  a=  25°  18'  45", 

A=  15°  58'  15". 

Ans.  Impossible  ;  why? 

25. 

Given  a=   32°   9'  17", 

b=   32°  41'; 

find  A=  49°  20'  17", 

B=  50°  19'  16", 

c=  44°  33'  17". 

26. 

Given  a=   55°  18', 

b=  39°27'; 

find  A=  66°  15'  6", 

B  =  45°   1'31", 

c=  63°55'21". 

27. 

Given  a=   56°  20', 

b=  78°  40'; 

find  A=  56°  51'   7", 

B=  80°31'48", 

c=  83°44'44i". 

28. 

Given  a=  86°  40', 

6=   32°  40'; 

find  A=  88°11'57".8, 

B=  32°42'37".8, 

c=  87°11'39".8. 

29. 

Given  a=   37°  48'  12", 

b=  59°44'16"; 

find  A=  41°  55'  45", 

B=   70°  19'  15", 

c=  66°  32'  6". 

30. 

Given  a=116°, 

b=   16°; 

find  A=   97°39'24".4, 

B=  17°41'39".9, 

c=114°55'20".4. 

31. 

Given  A  =  52°  26', 

B=  49°  15'; 

find   o=  36°24'34".5, 

b=  34°33'40", 

c=  48°  29'  20". 

32. 

Given  A  =   64°  15', 

B=  48°  24'; 

find   a=  54°  28'  53", 

b=  42°30'47", 

c=  64°  38'  38". 

33. 

Given  A  =  54°   1'15", 

B  =  137°  24'  21"; 

find   a=   29°  46'  8", 

b  =  155°  27'  55", 

c=142°   9'  12". 

34. 

Given  A=  46°  59'  42", 

B=  57°  59'  17"; 

find   a=  36°  27', 

6=  43°32'37", 

c=  54°  20'  3". 

35. 

Given  A  =  55°  32'  45", 

B  =  101°  47'  56"; 

find   a=  54°  41'  35", 

6=104°21'28", 

c=  98°  14'  24". 

36. 

GivenA=   60°27'24".3, 

B=  57°16'20".2; 

find   o=  54°32'32".l, 

6=  51°43'36".l, 

c=   68°  56'  28".  9. 

EXAMPLES. 


319 


Solve  the  following  quadrantal  triangles  : 

37. 

Given  B=   74°  45', 
find    6=  85°17'15".5, 

a=  18°  12', 
A=   17°  34'  2", 

c=   90°; 
0  =  104°  31'  13". 

38. 

Given  A  =  110°  47'  50", 
find   a  =104°  53'   0".S, 

B  =  135°35'34''.5, 
6=133°  39'  47".7, 

c=   90°; 
C  =  104°41'37".2. 

Solve  the  following  oblique  triangles  : 

39. 

Given  a=   73°  58', 
find  A=116°   8'  28", 

6=  38°  45', 
B=  35°  46'  39", 

0=  46°  33'  39"; 
c=  51°   I'll". 

40. 

Given  a=  96°  24'  30", 
find  A=  97°53'Qi", 

6=  68°27'26", 
B=  67°59'39i", 

0=   84°  46'  40"; 
c=   87°31'37". 

41. 

Given  a=  76°  24'  40", 
find  A=  63C48'35V, 

6=  58°  18'  36", 
B=  51°46'12i", 

0  =  116°  30'  28"; 
c=104°13'27". 

42. 

Given  a=  86°18'40", 
find  A=  64°48'53|", 

6=  45°36'20", 
B=  40°23'15|", 

0  =  120°  46'  30"; 
c=  108°  39'  111". 

43. 

Given  a=  88°  24', 
find  A=  65°  13'  3%", 

6=  56°  48', 
B=  49°  27'  51", 

0  =  128°  16'; 
c=  120°  10'  52". 

44. 

Given  a=  68°  20'  25", 
find  A=  56°  16'  15", 

6=  52°  18'  15", 
B=  45°  4'  41", 

0  =  117°  12'20"; 
c=  96°20'44". 

45. 

Given  a=  88°  12'  20", 
find  A=   63°  15'  12", 

6=124°   7'17", 
B  =  132°  17'  59", 

0=  50°  2'  1"; 
c=  59°  4'25". 

46. 

Given  a=  32°23'57", 
find  A=  60°  53'  2", 

6=  32°23'57", 
B=  60°  53'  2", 

0=   66°  49'  17"; 
c=  34°  19'  11". 

47. 

Given  6=  99°  40'  48", 
find  B=  95°  38'  4", 

c=100°49'30", 
0=  97°26'29".l, 

A=   65°33'10"; 
a=  64°23'15".l. 

48. 

GivenA=  31°34'26", 
find  a=  40°  1'  5J", 

B=  30°  28'  12", 
6=  38°31'  3J", 

c=  70°   2'  3"; 
0  =  130°   3'  50". 

320 


SPHERICAL    TRIGONOMETRY. 


GivenA  =  130°5'22".4, 

B=  32°26'6".41, 

c=   51°   6'11".6; 

find   a  =   84°  14'  29", 

6=   44°  13'  45", 

C^  36°  45'  26". 

Given  A  =   96°  46'  30", 

B=   84°  30'  20", 

c=126°46'; 

find   a=102°21'42", 

6=   78°  17'   2", 

C  -125°  28'  13  J". 

Given  A  =  84°  30'  20", 

B=   76°  20'  40", 

c=130°46'; 

find   a=   94°  34'  52  J", 

b=   76°  40'  48i", 

C  =  130°  51'  33|". 

Given  A=  107°  47'   7", 

B=  38°58'27", 

c=   51°  41'  14"; 

find   a=   70°  20'  50", 

b=  38°  27'  59", 

C=  52°  29'  45". 

GivenA  =  128°41'49", 

B  =  107°  33'  20", 

c=124°12'31"; 

find   a=125°44'44", 

b=  82°47'35", 

C  =  127°22'   7". 

GivenA=129°58'30", 

B=  34°  29'  30", 

c=   50°   6'20"; 

find   a=  85°  59', 

b=  47°29'20", 

C=  36°   6'  50". 

Given  A=   95°  38'  4", 

C=  97°26'29", 

6=   64°  23'  15"; 

find   a=   99°  40'  48", 

c=  100°  49'  30", 

B=   65°  33'  10". 

Given  A  =   70°, 

B  =  131°18', 

c—1160; 

find   a=  57°  56'  53", 

b  =  137°  20'  33", 

C=   94°48'12". 

Given  a=   62°  15'  24", 

&  =  103°18'47", 

A=   53°  42'  38"; 

find  B=  62°24'24".8, 

C  =  155°43'll".3, 

c=153°  9'  35J", 

or     B'=117°35'35".2, 

Cf=  59°   6'10".6, 

c'=  70°25'26". 

Given  a=  52°  45'  20", 

b=  71°12'40", 

A=  46°  22'  10"; 

find  B=  59°  24'  15  1", 

C  =  115°  39'  55J", 

c=  97°33'18".8, 

or     B'=120°35'44i", 

C'=  26°59'55".2, 

c'=  29°57'10".5. 

Given  a=  48°45'40", 

b=  67°12'20", 

A=  42°  20'  30"; 

find  B=  55°  39'  57", 

0  =  116°  34'  18", 

c=   93°   8'   9".6, 

or     B'=124°20'   3", 

C'=  24°  32'  15", 

c'=  27°  37'  20". 

Given  a=  46°  20'  45", 

b=  65°  18'  15", 

A=   40°  10'  30"; 

find  B=   54°   6'  19",       C  =  116°  55'  26", 

c=  90°  31'  46", 

or     B'  =  125°53'41",     ;  C'=   24°12'53".3, 

c'=  27°  23'  14". 

EXAMPLES. 


321 


61. 

Given  a=150°57'  5", 

6  =  134°  15'  54", 

A  =  144°22'42"; 

find  B  =  120°47'44", 

0=  97°  42'  55", 

c=  55°  42'  8", 

or     B'=  59°  12'  16", 

C'=  29°  9'  9", 

c'=  23°57'29". 

62. 

Given  a=  50°45'20", 

b=  69°12'40", 

A=   44°  22'  10"; 

find  B=  57°34'51".4, 

C  =  115°57'50".6, 

c=  95°18'i^".4, 

or     B'=122°25'  8".6, 

C'=  25°44'31".6, 

c'=  28°  45'  5".2. 

63. 

Given  a=  40°   5'25".6, 

6==118°22'   7".3, 

A=  29°42'33".8; 

find  B=  42°37'17".5, 

C  =  160°   1'24".4, 

c=153°38'42".4, 

or     B'=137°22'42".5, 

C'=  50°18'55".2, 

c'=  90°   5'41".0. 

64. 

Given  a=   99°40'48", 

b=  64°23'15", 

A=  95°  38'   4"; 

find  B=   65°  33'  10", 

0=  97°26'29", 

c=  100°  49'  30". 

(No  ambiguity  ;  why?) 

65. 

Given  A  =   79°  30'  45", 

B=  46°  15'  15", 

a=   53°  18'  20"; 

find    b=  36°   5'34f", 

c=  50°24'57", 

C=  70°  55'  35". 

(No  ambiguity  ;  why?) 

66. 

Given  A  =  73°  11'  18", 

B  =  61°  18'  12", 

a=  46°  45'  30"; 

find    b=  41°52'34f", 

c=  41°  35'  4", 

C=  60°42'46".5. 

(Only  one  solution  ;  why?) 

57. 

GivenA=  46°30'40", 

B=  36°  20'  20", 

o=   42°15'20"; 

find   b=  33°18'47£", 

c=   60°  32'   6", 

0  =  110°   3'14".6. 

(Only  one  solution  ;  why?) 

68. 

Given  A  =   61°  29'  30", 

B=  24°  30'  30", 

a=   34°  30"; 

find    b=   15°30'30".5, 

c=  39°  33'  52", 

0=  98"48'58".5. 

(Only  one  solution  ;   why?) 

69. 

Given  A  =  36°  20'  20", 

B=  46°30'40", 

a=  42°  15'  20"; 

find    b=  55°  25'   2f  , 

c=  81°27'26J", 

C  =  119°22'27i", 

or      &'=124°34'57i", 

c'=162°34'27", 

C'=164°41'55". 

322                           SPHERICAL   TRIGONOMETRY. 

70. 

Given  A  =  52°  50'  20", 

B  =  66°   7'  20", 

a=   59°  28'  27"; 

find    b=  81°  15'  15", 

c=110°10'50|", 

C  =  119°  43'  48", 

or      b'=   98°  44'  45", 

c'=138°45'26", 

C'=142°24'59". 

71. 

Given  A  =  115°  36'  45", 

B=  80°  19'  12", 

b=  84°21'56"; 

*  find   a  =  114°  26'  50", 

c=  82°  33'  31", 

C=  79°  10'  30". 

72. 

Given  A=   61°  37'  52".  7, 

B  =  139°54'34".4, 

6  =  150°17'26".2; 

find   a  =  42°37'17".5, 

c=129°41'  4".8, 

C=   89°54'19".0, 

or      a'=137°22'42".5, 

cf=  19°58'35".6, 

C'=  26°21'17".6. 

73. 

Given  A  =   70°, 

B  =  120°, 

b=  80°. 

Ans.  Impossible  ;  why  ? 

74. 

Given  a  =108°  14', 

b=  75°  29', 

c=   56°  37'; 

find  A=123°53'47", 

B=  57°  46'  56", 

C=  46°51'51".5. 

75. 

Given  a=   57°  17', 

b=  20°39', 

c=   76°  22'; 

find  A  =  21°   1'   2", 

B=     8°  38'  46", 

C  =  155°31'36".5. 

76. 

Given  a=  68°  45', 

b=  53°  15', 

c=  46°  30'; 

find  A=   94°  52'  40", 

B=  58°   5'  10", 

C=  50°50'52i". 

77. 

Given  a  =   63°  54', 

b=  47°  18', 

c=  53°  26'; 

find  A  =  86°  30'  40", 

B=  54°  46'  14", 

C=  63°12'55i". 

78. 

Given  a=  70°  14'  20", 

b=  49°  24'  10", 

c=  38°  46'  10"; 

find  A=  110°  51  '16", 

B=  48°  56'  4", 

C=  38°  26'  48". 

79. 

Given  a=124°12'31", 

6=  54°  18'  16", 

c=   97°12'25"; 

find  A  =  127°  22'   7", 

B=  51°18'11", 

0=  72°  26'  40". 

80. 

Given  a  =   50°  12'  4", 

6=116°  44'48", 

c=129°ll'42"; 

find  A=   59°  4'  25", 

B=  94°  23'  10", 

0  =  120°  4'  50". 

81. 

Given  a  =100°, 

b=  50°, 

c=  60°; 

find  A  =  138°15'45".4, 

B=  31°11'14".0, 

C=  35°49'58".2. 

82. 

Given  A  =   86°  20',             B=   76°  30', 

C=   94°  40'; 

find   a=  87°  20'  28", 

6=  76°  44'  2V, 

c=  93°  55'  31". 

EXAMPLES. 


323 


83. 


84. 


85. 


86. 


87. 


89. 


Given  A  =  96°  45', 

find   a=  88°  27' 49", 

Given  A  =   78°  30', 
find    a=   74°57'46", 

Given  A  =   57°  50', 
find   a=  58°   8' 19", 

Given  A  =  129°   5' 28", 
find   a  =  135°  49' 20", 

Given  A  =  138°  15' 50", 
find   a=100°   0'   8".4, 

GivenA=102°14'12", 
find   a=104°25'   8", 

Given  A=  20°   9' 56", 
find   a=   20°  16' 38", 


B  =  108°30', 

b  =  107°  19'  52", 

B  =  118°40', 

6=120°   8'49", 

B=  98°  20', 
b=  83°   5' 36", 

B  =  142°  12'  42", 
6=144°37'15", 

b=  49°  59' 56".  4, 

B=  54°  32' 24", 
b=  53°  49' 25", 

B=  55°  52' 32", 


c=115°28'13j". 

C=  93°  20'; 
c=100°18'llf". 

C=  63°  40'; 
c=   64°   3' 20". 

C  =  105°   8' 10"; 
c=   60°   4' 54". 

C=  35°  50'; 
c=  60°   0'11".2. 

C=  89°   5' 46"; 
c=  97°  44' 18". 

C  =  114°  20'  14"; 


b=  56°19'41",     i  c=  66°20'43". 


90.    If  a,  6,  c  are  each  <  ^,  show  that  the  greater  angle  may 

u 


exceed  - 


91.  If  a  alone  >  ^TT,  show  that  A  must  exceed  -• 

2 

92.  If  a  and  b  are  each  >^-TT,  and  c  <frj7r,  prove  that : 

(1)  The  greatest  angle  A  must  be  >  £TT  ; 

(2)  B  ma?/  be  >  £TT; 

(3)  C  may  or  may  not  be  <  £TT. 

93.  If  cos  a,   cos  6,   cos  c   are  all   negative,  prove  that   cos  A, 
cos  B,  cos  C  are  all  necessarily  negative. 

94.  In  a  spherical  triangle,  of  the  five  products,  cos  a  cos  A, 
cos  b  cos  B,  cos  c  cos  C,  cos  a  cos  b  cos  c,   —  cos  A  cos  B  cos  C,  show 
that  one  is  negative,  the  other  four  being  positive. 


324 


SPHERICAL    TRIGONOMETRY. 


CHAPTER   XII. 
THE  IN-OIEOLES  AND  EX-OIEOLES,  —  AEEAS, 

215.   The    Ill-Circle    (Inscribed    Circle).  —  To  find    the 
angular  radius  of  the  in-circle  of  a  triangle. 

Let  ABC  be  the  triangle  ;  bisect  the  angles  A  and  B 
by  the  arcs  AO,  BO;  from  O  draw 
OD,  OE,  OF  perpendicular  to  the 
sides.  Then  it  may  be  shown  that 
0  is  the  in-centre,  and  that  the  per- 
pendiculars OD,  OE,  OF  are  each 
equal  to  the  required  angular  radius. 

Let  2  s  =  the  sum  of  the  sides  of 
the  triangle  ABC.  The  right  triangles 
OAE,  OAF  are  equal. 

.-.  AF  =  AE. 


Similarly, 


BD  =  BF,  and  CD  =  CE. 


Now  tan  OF  =  tan  OAF  sin  AF    .     (Art.  186) 

or,  denoting  the  radius  OF  by  ?•,  we  have 

j^ 

tan  r  =  tan  —  sin  (s  —  a) (1) 


or     tan  r  =  y/sin  (*  ~  a^  sin  *.*  ~  6)  si"  ^  ~  °^ 
Af  sins 


n 
sins 


(Art.  195)  (2) 


THE  ESCRIBED   CIRCLES.  325 

Also,  sin(s—  a) 


c)  cos^a  —  cos  £(&  +  c)  sinja 

(Art.  198) 


sin- 

_  sin  a  sin^B  sin-|-C 

sin^-A 
which  in  (1)  gives 

B    .    C 

sin  —  sin  — 

o         o 

tan  r  =  -  =—  ?  —  sin  a      .........     (3) 

cos^A 

.     .     (Art.  196)  (4) 


2  cos  ^  A  cos  £  B  cos  ^  C 
an  equation  which  is  equivalent  to  the  following  : 

cotr  =  ^-[cosS+cos(S-A)+cos(S-B)+cos(S-C)](5) 

216.  The  Ex-Circles.  —  To  find  the  angular  radii  of  the 
ex-circles  of  a  triangle. 

A  circle  which  touches  one  side  of  a  triangle  and  the 
other  two  sides  produced,  is  called  an  escribed  circle,  or 
ex-circle,  of  the  triangle.  It  is  clear  that  the  three  ex-circles 
of  any  triangle  are  the  in-circles  of  its  colunar  triangles 
(Art.  191,  Sch.). 

Since  the  circle  escribed  to  the  side  a  of  the  triangle 
ABC  is  the  in-circle  of  the  colunar  triangle  A'BC,  the  parts 
of  which  are  a,  TT  —  6,  TT  —  c, 
A,  TT  —  B,  TT  —  C,  the  problem 
becomes  identical  with  that  A< 
of  Art.  215 ;  and  we  obtain 
the  value  for  the  in-radins  of 
the  colunar  triangle  A'BC,  by  substituting  for  6,  c,  B,  C, 
their  supplements  in  the  five  equations  of  that  article. 


326  SPHERICAL    TRIGONOMETRY. 

Hence,  denoting  the  radius  by  ra,  we  get 

tan  ra  =  tan  i  A  sin  s (1) 

(2) 


sin(s  —  a) 

==cosj_    >S]L_gina     .     .     .     . 

cos^- A 

2cos-i-AsiniB  sin-|C 
cotra=^-[— cosS  — cos(S  — A)-f-cos(S— B)  +  cos(S 

These  formulae  may  also  be  found  independently  by 
methods  similar  to  those  employed  in  Art.  215,  for  the 
in-circle,  as  the  student  may  show. 

tSch.  Similarly,  another  triangle  may  be  formed  by  pro- 
ducing BC,  BA  to  meet  again,  and  another  by  producing 
CA,  CB  to  meet  again.  The  colunar  triangles  on  the  sides 
b  and  c  have  each  two  parts,  b  and  B,  c  and  C,  equal  to 
parts  of  the  primitive  triangle,  while  their  remaining  parts 
are  the  supplements  in  the  former  case  of  a,  c,  A,  C,  and  in 
the  latter,  of  a,  6,  A,  B. 

The  values  for  the  radii  rb  and  rc  are  therefore  found  in 
the  same  way  as  the  above  values  for  ra\  or  they  may  be 
obtained  from  the  values  of  ra  by  advancing  the  letters. 

Thus.  tan  rb  =*  tan  4-  B  sin  s  = ,  etc., 

sm(s  —  b) 

and  tan  rc  =  tan  i  C  sin  s  =   .     n — -,  etc. 

sm(s  —  c) 

217.  The  Circumcircle.  —  To  find  the  angular  radius  of 
the  circumcircle  of  a  triangle. 

The  small  circle  passing  through  the  vertices  of  a  spheri- 
cal triangle  is  called  the  circumscribing  circle,  or  circumcircle, 
of  the  triangle. 


THE  CIECUMCIECLE. 


327 


Let  ABC  be  the  triangle;  bisect  the 
sides  CB,  CA  at  D,  E,  and  let  0  be  the 
intersection  of  perpendiculars  to  CB, 
CA,  at  D,  E;  then  0  is  the  circimi- 
centre. 

For,  join  OA,  OB,  OC ;  then  (Art.  186) 

cos  OB  =  cos  BD  cos  OD, 
cos  OC  =  cos  DC  cos  OD. 
.-.  OB  =  OC.     Similarly,  OC  =  OA. 
Now  the  angle 

OAB  =  OB  A,  OBC  =  OCB,  OCA  =  OAC 


Let  OC  =  R ;  then,  in  the  triangle  ODC,  we  have 

cos  OCD  =  tan  CD  cot  CO  =  tan  1  a  cot  R  .     (Art.  186) 
tan  i  a 


tanR 


or 


tan  R  =  — 


cos(S- A) 
cos  S 


.     .     .     .     (1) 
(Art.  196)  (2) 


Also  cos(S  - A)=  cos  i[(B  +  C)  -  A] 
=  cosi(B  +  C)  cos|  A  +  sin|(B  +  C)  si: 


-c)]  (Art.  198) 


sin  A 


cos  ^6  cos  |  c, 


cosset 
which  in  (1)  gives 
tanR  = 


sin-^q 


sin  A  cos  ^b  cos ! c 
_  2  sin  ±  a  sin  |  b  sin  j 


.    .    .    .     (3) 
(Art.  195)  (4) 


328  SPHERICAL   TRIGONOMETRY. 

which  may  be  reduced  to  the  following  : 

tan  R  =  —  [sin(s  —  a)  +  sin(s  —  b)  +  sin(s  —  c)  —  sins]    (5) 

—  ft 

218.  Circumcircles  of  Colunar  Triangles.  —  To  find  the 
angular  radii  of  the  circumcircles  of  the  three  colunar  triangles. 

Let  KU  R2,  R3  be  the  angular  radii  of  the  circumcircles  of 
the  colunar  triangles  on  the  sides  a,  b,  c,  respectively. 
Then,  since  Rj  is  the  circumradius  of  the  triangle  A'BC 
whose  parts  are  a,  TT  —  6,  TT  —  c,  A,  TT  —  B,  TT  —  C,  we  have, 
from  Art.  217, 


(2) 


,r,  -p  sin  A  a  /Q\ 

tanK1==-  —  —  -  —  f—  —  —  —    ........     (3) 

sin  A  sin    b  sin    c 


(4) 


tan  Rj  =  —  [sins—  sin(s—  a)+sin(s—  6)+sin(s—  c)]  (5) 

^  n 

Similarly, 

-P  tan-i-6      cos(S  —  B) 

tan  K2  =  ---  ^—  =  -  i  —    —  '-  =  etc., 
cosS  N 

and  tanR3  =  -^n^=cos(S-C)=etc. 

cosS  N 

EXAMPLES. 

Prove  the  following  : 

1.  cos  s  +  cos  (s  —  a)  -f  cos  (s  —  b)  -f-  cos  (s  —  c) 

=  4  cos  \  a  cos  \  b  cos  J  c. 

2.  cos  (s  —  6)  -f  cos  (s  —  c)  —  cos  (s  —  a}  —  cos  s 

=  4  cos  J  a  sin  £  6  sin  -J-  c. 


PROBLEM.  329 

3. 


cosJB  2  cos  ^-B  sin  ^C  sin  J  A 


cos  4-  A  cos  4-  B    . 
4.   tan  rc  =  —  — %—  sin  c  = 


cos-J-0  2cos£Csin^-Asin^-B 

5.  cot  r  :  cot  ^  :  cot  r2  :  cot  r3 

=  sins  :  sin(s  —  a)  :  sin(s  —  6)  :  sin(s  —  c). 

6.  tan  r  tan  i\  tan  r2  tan  r3  =  n2. 

1.   cot  r  tan  r±  tan  ?'2  tan  r3  =  sin2  s. 

8. 


q    ,      T>  _  2  cos  £  a  cos  J  b  sin  J  c 
71 

10.  tanE1:tanE2:tanK3=cos(S-A):cos(S-B):cos(S-C). 

11.  cot  E  cot  E!  cot  E2  cot  E3  =  N2. 

12.  tan  E  cot  Ej  cot  E2  cot  E3  =  cos2  S. 

AREAS    OF    TRIANGLES. 

219.  Problem.  —  To  find  the  area  of  a  spherical  triangle, 
having  given  the  three  angles. 

Let   r  =  the  radius  of  the  sphere. 

E  =  the  spherical  excess  =  A  +  B  +  C  -  180°. 
K  =  area  of  triangle  ABC. 

It  is  shown  in  Geometry  (Art.  738)  that  the  absolute  area 
of  a  spherical  triangle  is  to  that  of  the  surface  of  the  sphere 
as  its  spherical  excess,  in  degrees,  is  to  720°. 


330  SPHERICAL    TRIGONOMETRY. 

Cor.    The  areas  of  the  colunar  triangles  are 


° 


180°  180°  180° 

220.   Problem.  —  To  find  the  area  of  a  triangle,  having 
given  the  three  sides. 

Here  the  object  is  to  express  E  in  terms  of  the  sides. 

I.    CagnolCs  Theorem. 
sin  |E  =  sin  £  (  A  +  B  +  C  -  TT) 
=  sin  J(A  -f  B)  sin  $  C  —  cos  £(A  +  B)  cos  JC 

=  sin  fr  C  cos  |C  pcQs  ^,    _  ^  _  GQg  ^q  +  ^  j        (Art.  198) 
cos^c 

sin  1  a  sin  4  6  sin  C      sin  £  a  sin  -J  6         2w        /A   ,   iar\si\ 
—  -  -  —  --  —  --  •  —  :  -  :  —  -  l  Art.  iyo  )  (^i  ) 
cos^c  cos^-c        sin  a  sin  o 

(2) 


2  cos  £  a  cos  J  6  cos  £  c 
II.    Lhuilier's  Theorem. 


sin  j(A  +  B)  -  sin  ^(TT  -  C) 
cos  i(A  +  B)  +  cos  |(TT  -  C) 

sin  \  (  A  +  B)  —  cos  \  G 


a  —  6)  —  cos-^c  ^  cos^-C 
+  cos^c    sin-JC 


cos  Js  cos  ^(s  —  c) 


(  Art 


=  Vtan|stanJ(s-a)tan^(s-6)tani-(s-c)    (Art.  195)  (3) 


AREAS   OF  TRIANGLES. 


331 


221.   Problem.  —  To  find  the  area  of  a  triangle,  having 
given  two  sides  and  the  included  angle. 

cos^E^  cos  [£  (A  +  B)  -  (iTr  -  iC)] 

B)  cos^C 

os2|-C    (Art.  198) 

]sec|-c    .     .     (1) 

Dividing  (1)  of  Art.  220  by  this  equation,  and  reducing, 
we  have 


t      iF— 


tan  %  a  tan  ^  b  cos  0 


EXAMPLES. 


(2) 


1.  Given  a  =  113°2'56".64,  b  =  82°  39'  28". 4,  c  =  74°54f 
31".06;  find  the  area  of  the  triangle,  the  radius  of  the 
sphere  being  r. 

By  formula  (3)  of  Art.  220, 
a  =  113°  2' 56 ".64 


b=    82°  39' 28  ".40 
c=    74°  54' 31  ".06 

2s  =  270°36'56".10 

s  =  135°18'28".05 

s-a=    22°  15' 31  ".41, 

s-b=    52°  38' 59".  65, 

s-c=    60°23'56".99. 


is  =  67°39'14".025 
-a)  =  ll°  7'45".705 
-6)  =  26°19'29".825 
_c  =  30°ll'58".495 


log  tan  |s  =  0.3860840 
log  tan  |0  -  a)  =  9.2938583 
log  tan  l(s  -b)  =  9.6944058 
log  tan  i(«  -  c)  =  9.7649261 

log  tan2iE  =  9.1392742 
log  tan  IE  =  9.5696371. 

1E  =  20°21'58".25. 
E  =  81°  27'  53" 

=  293273". 
7T?-2   .     .     .     .     [(1)  of  Art.  219] 


332  SPHERICAL    TRIGONOMETRY. 

2.  Given  A  =  84°  20'  19",  B  =  27°  22'  40",  C  =  75°33'; 
find  E  =  7°15'59". 

3.  Given  a  =  46°  24',  b  =  67°  14',  c  =  81°12'; 
find  K  =  ii 


4.  Given  a  =  108°  14',  b  =  75°  29',  c  =  56°  37'  ; 
find  E  =  48°  32'  34".5. 

5.  Prove  cosjE  =  1  +  cosa  +  cos6  +  cosc 

4cosiacos|6cosic 

_  cos2ig  -f  cos2i&  -f  cos2ic  — 


cos^c 


6.  "  sin  *  E=    /si 

\ 

7.  ucos1E=    /CQ 

\ 


sinC 

_  cot|-6  cot^c  +  cos  A 
sin  A 

cot  ^  c  cot  ^  a  -\-  cos  B 
sinB 


EXAMPLES. 

Prove  the  following : 

1.  sin  (s  —  a)  -f  sin  (s  —  b)  +  sin  (s  —  c)  —  sin  s 

=  4  sin  la  sin  J6  sin^c. 

2.  sin  s  -f  sin  (s  —  6)  +  sin  (s  —  c)  —  sin  (.9  —  a) 

=  4  sin  i  a  cos  ^  6  cos  ^  c. 


EXAMPLES.  333 

3.  sin  (s  —  b)  sin  (s  —  c)  4-  sin  (s  —  c)  sin  (s  —  a) 

4-  sin  (s  —  a)  sin  (s  —  b)  4-  sin  s  sin  (s  —  a) 
4-  sins  sin(s  —  b)  4-  sins  sin(s  —  c) 
=  sin  b  sin  c  4-  sin  c  sin  a  4-  sin  a  sin  b. 

4.  sin  (s  —  b)  sin  (s  —  c)  4-  sin  (s  —  c)  sin  (s  —  a) 

—  sin(s  —  a)sin(.s  —  b)  4-  sins  sin(s— a) 
-f  sin  s  sin  (s  —  b)  —  sin  s  sin  (s  —  c) 
=  sin  6  sin  c  4-  sin  c  sin  a  —  sin  a  sin  b. 

5.  sin2s  +  sin2(s  —  a)  4-  sirr(s  —  b)  +  sin2(s  —  c) 

=  2(1  —  cos  a  cos  6  cos  c) . 

6.  sin2s  4-  sin2(s  —  a)  —  sin2(s  —  6)  —  sin2(s  —  c) 

=  2  cos  a  sin  6  sin  c. 

7.  cos2s  4-  cos2(s  —  a)  4-  cos2(s  —  5)  4-  cos2(s  —  c) 

=  2  (1  4-  cos  a-  cos  b  cos  c) . 

8.  cos2s  4-  cos2(s  —  a)  —  cos2(*  —  b)  —  cos2(s  —  c) 

=  —  2  cos  a  sin  b  sin  c. 

9.  tan  r  cot  ^  tan  r2  tan  r3  =  sin2  (s  —  a) . 

10.  tanr  tanrx  cotr2  tanr3  =  sin2(s  —  6). 

11.  tan  r  tan  r±  tan  r2  cot  ?*3  =  sin2(s  —  c) . 

12.  cotr  sins  =  cot^-Acot^-B  cot^C. 

13.  tan  T-J  -h  tan  r2  4-  tan  r3  —  tan  r  =  — 

14.  cot  7'j  4-  cot  r2  4-  cot  ?:3  —  cot  r  = 

15.  tan  r, 


sin  A  sin  B  sin  C 
4  sin  ^  a  sin  ^  b  sin  £  c 

sin  b  sin  c 


1  +  cos  A  1  +  cos  B  1  +  cos  C 

-i  -  tan  7%  4-  tan  r2  4-  tan  r« — tan  r 

16.  -  =|(l4-cosa4-cos&4-cosc). 

cot  TI  4-  cot  r2  4-  cot  rB — cot  r 


334  SPHERICAL    TRIGONOMETRY. 

17.    cot'r,  +  cot2,-,  +  cot'r,  +  cotV  =  gjl-eosacogftcogc). 

n2 

18        1     _i_      1  1  1      _  2  cos  a  sin  b  sin  c 

sin2r      sin2?^      sin2r2      sin27*3  n2 

19.  cot  r2  cot  r3  4-  cot  rs  cot  7\  +  cot  r±  cot  7*2 

+  cot  7*  (cot  ?*!  4-  cot  r2  +  cot  r3) 
_  sin  6  sin  c  -f-  sin  c  sin  a  4-  sin  a  sin  6 
7r 

20.  tan  ?*2  tan  ?*3  +  tan  7*3  tan  ^  +  tan  ^  tan  r2 

4-  tan  r(tan  7^  4-  tan  r»  4-  tan  rs) 
=  sin  b  sin  c  +  sin  c  sin  a  4-  sin  a  sin  &. 

21.  cot  R  tan  E-!  cot  B2  cot  K3  =  cos2  (S  —  A) . 

22.  cot  E  cot  R!  tan  E2  cot  R3  =  cos2  (S  -  B) . 

23.  cot  E  cot  E!  cot  E2  tan  E3  =  cos2  ( S  —  G ) . 

24.  tan  Ex  4-  tan  E2  =  cot  r  +  cot  r3. 

25.  tan  Ex  4-  tan  E2  4-  tan  E3  —  tan  E  =  2  cot  r. 

26.  tan  E  —  tan  Ej  +  tan  E2  4-  tan  E3  =  2  cot  i\. 

27.  tan  E  4-  tan  Ej  —  tan  E2  4-  tan  E3  =  2  cot  7'2. 

28.  tan  E  4-  tan  Ej  4-  tan  E2  —  tan  E3  =  2  cot  rs. 

29.  cot  7*!  4-  c°t  ^2  +  c°t  ^*3  —  cot  r  =  2  tan  E. 

30.  cot  r  —  cot  7*j  4-  cot  r2  4-  cot  r3  =  2  tan  Ex. 

31.  cot  r  4-  cot  9'j  —  cot  r2  4-  cot  ?3  =  2  tan  E2. 

32.  cot  r  4-  cot  rx  -f  cot  r2  —  cot  ?*8  =  2  tan  E3. 

33.  tan  E  4-  cot  r  =  tan  E,  4-  cot  r,  =  etc., 

=  £  (cot  r  4-  cot  TI  4  cot  r2  +  cot  r3) . 

34.  tan2E  +  tan2E14-tan2E24-tan2E3 

_  2(1  +  cos  A  cos  B  cos  C) 

~~ 


EXAMPLES.  335 

35     tan2  B  4-  tan2  B^  +  tan2  E2  +  tan2  E3  =  ^ 
cot2  r  4-  cot2  TI  4-  cot2  r2  +'  cot2  r3 

36.  tan2  E  +  tan2  Ej  -  tan2  E2  -  tan2  E3 

2  (cos  A  sin  B  sin  C) 

N2 

Orr  ,o         2  cos  a  sin  6  sin  c 

37.  eot2  r  4-  cot2 1\  —  cot2  r2  —  cot2  r3  =  —  — 

n2 

38     tan2  E  4-  tan2  E!  -  tan2  E2  -  tan2  E8  =._  cosA< 
cot2  r  +  cot2  rx  —  cot2  r2  —  cot2  r3  cos  a 

39.  tan  E  cot  Ej  =  tan  £  &  tan  ^  c. 

40.  (cot  r  +  tan  B)'+  1  =  ( gin  ffi  +  si"  &  +  sin  CY. 

\  2n  y 

-r,  \  o  ,          /sin  6  +  sin  c  —  sin  a\2 

41.  (cot  rj  —  tan  E)2+  1  =  ( -  )  • 

\  2n  J 

N 

42.  tan  ^  A  sin  (s  —  a)  = 


2  cos  J  A  cos  ^B  cos  ^  C 

jo     tanr  _cos(S  —  A)cos(S  —  B)  cos(S  — C) 
tanE  2  cos  JAcos^B  cos^C 

44.  cot  (s  —  b)  cot  (s  —  c)  -f-  cot  (s  —  c)  cot  (s  —  a) 

4-  cot(s  —  a)  cot(s  —  6)  =  cosec2r. 

45.  eot  (s  —  b)  cot  (s — c)  —  cot  s  cot  (s — b)  —  cot  s  cot  (s — c) 

=  cosec27y 

46.  cot  (s — c)  cot  (s — a)  —  cot  s  cot  (s — c)  —  cot  s  cot  (s — a) 


=  cosec2r2. 


47.    cot(s— a)cot(s— b)  —  cot  s  cot  (s— a)  —  cot  s  cot  (s  —  6) 

=  cosec2r3. 

^o     cot(g  — a)   ,  cot(s  —  6)   ,  cot(s  —  c)   ,  2 cots 

^"*     : — ^ 1 :;    *  r         .    „  i       .    „ 

sjirrj  sm2r2  sm2r3  sm2r 

=  3cot(s  —  a)  cot(«  —  6)cot(s  —  c). 


336  SPHERICAL    TRIGONOMETRY. 

49.  cosec2 1\  -f  cosec2  r2  -f-  cosec2  r3  —  cosec2  r 

=  —  2  cot  s[cot(s  —  a)  +  cot(s  —  b)  +  cot(s  —  c)]. 

50.  -1-  +  -J--+      *      +./ 


sm2r2      sin2r3 


-22  tan  (s  -  a)  tan  (s  -  b) 
tan  s  tan  (s  —  a)  tan  (s  —  b)  tan  (s  —  c) 

2  N  cos  s 


51.  cot  B.  —  cot  Bj  —  cot  B2  —  cot  B3  = 

n 

52.  sin(A-£E)  = ™ 

2  cos  -j-  a  sin  -j-  6  sin  ^  c 

53.  sin(B-*E)  =  -  n^   .        . 

J  sin  ^  a  cos  f  o  sin  ^  c 


54.    sin(C-iE)  =  — 


2  sin  £  a,  sin  J  6  cos  -J  c 

t-t-  /A        1  TT<\      sin2  4-  6  -f-  sin2  4-  c  —  sin24-a 

55.    cos(A  — ^E)=—  —2 — 

2  cos  -J-  a  sin  -J-  6  sin  J  c 


56.    cos(B  -  JE)  = 

57. 

58. 


sinC 
_  tan  ^  6  tan  ^c  +  cos  A  _  tan  |q  cot  ^6  —  cosB 


sin  A  sin  B 

59. 

60. 
61. 


62.  If  S,  Sj,  S2,  S3  denote  the  sums  of  the  angles  of  a  triangle 
and  its  three  colunars,  prove  that  S  -j-  Sj  +  ^2  +  S3  =  3  IT. 

63.  In  an  equilateral  triangle,  tan  E  =  2  tan?*. 


EXAMPLES.  337 

64.  If  EU  E2,  E3  denote  the  spherical  excesses  of  the  cohmars 
on  a,  b,  c,  respectively,  show  that  E  -f  Ex  +  E2  +  E3  =  2ir  ; 
and  therefore  the  sum  of  the  areas  of  any  triangle  and  its 
colunars  is  half  the  area  of  the  sphere. 

65.  Given  a  =  108°  14',  b  =  75°  29',  c  =  56°37'; 

find  E  =  48°  32'  34".5. 

66.  Given  a  =  63°  54',  b  =  47°  18',  c  =  53°26'; 

find  E  =  24°  29'  49*-". 

67.  Given  a  =  69°  15'  6",  b  =  120°  42'  47",  c  =  159°  18'  33"; 

find  E  =  216°  40'  23". 

68.  Given  a  =  33°l'45",  b  =  155°  5'  18",  0  =  110°  10'; 

find  E  =  133°  48'  55". 

69.  Given  a  =  b  —  c  —  1°,  on  the  earth's  surface  ; 

find  E  =  27".21. 

70.  Given  a  =  b  =  c  —  60°,  on  a  sphere  of  6  inches  radius  ; 
find  the  area  of  the  triangle.         Ans.  19.845  square  inches. 

71.  If  a=6=-,  and  c=~t  prove  sin£E  =  J,  and  cosE  =  J. 

3  2 

72.  If  G  =  ~9   prove   sin  ^  E  =  sin  £  a  sin  \  b  sec  £  c,   and 

& 

cos^-E  =  cosiacos£&  sec^c. 

73.  If  a  =  b,  and  C  =  -,  prove  tan  E  =  \  tan  a  sec  a. 


74.  If  A-hB-f-C  =  27r,  prove  cos2  Ja+cos2^&  +  cosz-Jc=l. 

75.  If  a  +  6  =  TT,  prove  that  E  =  C  ;  and  if  E'  denote  the 
spherical  excess  of  the  polar  triangle,  prove  that 


=  sin  a  cos 


76.   Prove  Bin'jE  =  Vsi"  *E  si"  *E'  f1" 

cot  £  a  cot  %  b 


338  SPHERICAL    TRIGONOMETRY. 


CHAPTER    XIII. 
APPLICATIONS   OF  SPHERICAL  TRIGONOMETRY, 

SPHERICAL    ASTRONOMY. 

222.  Astronomical  Definitions, 

The  celestial  sphere  is  the  imaginary  concave  surface  of 
the  visible  heavens  in  which  all  the  heavenly  bodies  appear 
to  be  situated. 

The  sensible  horizon  of  a  place  is  the  circle  in  which  a 
plane  tangent  to  the  earth's  surface  at  the  place  meets 
the  celestial  sphere. 

The  rational  horizon  is  the  great  circle  in  which  a  plane 
through  the  centre  of  the  earth  parallel  to  the  sensible 
horizon  meets  the  celestial  sphere.  Because  the  radius 
of  the  celestial  sphere  is  so  great,  in  comparison  with  the 
radius  of  the  earth,  these  two  horizons  will  sensibly 
coincide,  and  form  a  great  circle  called  the  celestial  hori- 
zon. 

The  zenith  of  a  place  is  that  pole  of  the  horizon  which 
is  exactly  overhead ;  the  other  pole  of  the  horizon  directly 
underneath  is  called  the  nadir. 

Vertical  circles  are  great  circles  passing  through  the 
zenith  and  nadir.  The  two  principal  vertical  circles  are 
the  celestial  meridian  and  the  prime  vertical. 

The  celestial  meridian  of  a  place  is  the  great  circle  in 
which  the  plane  of  the  terrestrial  meridian  meets  the 
celestial  sphere ;  the  points  in  which  it  cuts  the  horizon 
are  called  the  north  and  south  points. 


SPHERICAL  ASTRONOMY.  339 

The  prime  vertical  is  the  vertical  circle  which  is  per- 
pendicular to  the  meridian;  the  points  in  which  it  cuts 
the  horizon  are  called  the  east  and  west  points. 

The  axis  of  the  earth  or  of  the  celestial  sphere  is  the 
imaginary  line  about  which  the  earth  rotates. 

The  celestial  equator,  or  equinoctial,  is  the  great  circle  in 
which  the  plane  of  the  earth's  equator  intersects  the 
celestial  sphere. 

The  poles  of  the  equinoctial  are  the  points  in  which  the 
axis  pierces  the  celestial  sphere. 

Hour  circles,  or  circles  of  declination,  are  great  circles 
passing  through  the  poles  of  the  equinoctial. 

The  ecliptic  is  a  great  circle  of  the  celestial  sphere,  and 
the  apparent  path  of  the  sun  due  to  the  real  motion  of 
the  earth  round  the  sun. 

The  equinoxes  are  the  points  in  which  the  ecliptic  cuts 
the  equinoctial.  There  are  two,  called  the  vernal  and  the 
autumnal  equinox,  which  the  sun  passes  on  March  20  and 
September  22. 

The  obliquity  of  the  ecliptic  is  the  angle  between  the 
planes  of  the  ecliptic  and  equator,  and  is  about  23°  27'.  ' 

Circles  of  latitude  are  great  circles  passing  through  the 
poles  of  the  ecliptic. 

223.  Spherical  Coordinates,  —  The  position  of  a  point 
on  the  celestial  sphere  may  be  denoted  by  any  one  of  three 
systems.  In  each  system  two  great  circles  are  taken  as 
standards  of  reference,  and  the  point  is  determined  by 
means  of  these  circles,  which  are  called  its  spherical 
coordinates,  as  follows : 

I.    The  horizon  and  the  celestial  meridian  of  the  place. 

The  azimuth  of  a  star  is  the  arc  of  the  horizon  inter- 
cepted between  the  south  point  and  the  vertical  circle, 


340  SPHERICAL    TRIGONOMETRY. 

passing  through  the  star;  it  is  generally  reckoned  from 
the  south  point  of  the  horizon  round  by  the  west,  from 
0°  to  360°. 

The  altitude  of  a  star  is  its  angular  distance  above  the 
horizon,  measured  on  a  vertical  circle.  The  complement 
of  the  altitude  is  called  the  zenith  distance. 

II.  The  equinoctial  and  the  hour  circle  through  the  vernal 
equinox. 

The  right  ascension  of  a  star  is  the  arc  of  the  equinoctial 
included  between  the  vernal  equinox  and  the  hour  circle 
passing  through  the  star;  it  is  reckoned  eastward  from 
0°  to  360°,  or  from  Oh  to  24h. 

The  angle  at  the  pole  between  the  hour  circle  of  the 
star  and  the  meridian  of  the  place  is  called  the  hour  angle 
of  the  star. 

The  declination  of  a  star  is  its  distance  from  the  equinoc- 
tial, measured  on  its  hour  circle ;  it  may  be  north  or  south, 
and  is  usually  reckoned  from  0°  to  90°.  It  corresponds  to 
terrestrial  latitude. 

The  polar  distance  of  a  star  is  its  distance  from  the  pole, 
and  is  the  complement  of  its  declination.  The  right  ascen- 
sion and  declination  of  celestial  bodies  are  given  in  nautical 
almanacs. 

III.  The  ecliptic  and  the  circle  of  latitude  through  the  vernal 
equinox. 

The  latitude  of  a  star  is  its  angular  distance  from  the 
ecliptic  measured  on  a  circle  of  latitude ;  it  may  be  north 
or  south,  and  is  reckoned  from  0°  to  90°. 

The  longitude  of  a  star  is  the  arc  of  the  ecliptic  inter- 
cepted between  the  vernal  equinox  and  the  circle  of  latitude 
passing  through  the  star. 


SPHERICAL  COORDINATES. 


341 


224.  Graphic  Representation  of  the  Spherical  Coordi- 
nates. —  The  figure  will 
serve  to  illustrate  the  pre- 
ceding definitions.  0  is 
the  earth,  PHP'R  is  the 
meridian,  P  the  north  pole, 
HR  the  horizon,  EQ  the 
equinoctial,  Z  the  zenith. 
Then,  of  a  place  whose 
zenith  is  Z,  QZ  is  the  ter- 
restrial latitude ;  and  since 

QZ  ==  PR, 
.-.  PR  =  the  latitude. 

But  PR  is  the  elevation  of  the  pole  above  the  horizon. 

Hence  the  elevation  of  the  pole  above  the  horizon  is  equal  to 
the  latitude. 

Let  V  be  the  vernal  equinox,  and  let  S  be  any  heavenly 
body,  such  as  the  sun  or  a  star ;  then  its  position  is  denoted 
as  follows : 

VK  =  right  ascension  of  the  body  =  a, 

KS  •=  decimation  "    "       "  =  8, 

ZPS  or  QK  =  hour  angle      "     "       "  =  *, 

PS  =  north  polar  distance  "     "       "  =p, 

HT  =  azimuth  "     "       "  =  a, 

TS  =  altitude  "     "       «  =h, 

ZS  =  zenith  distance  "     "       "  =  z, 

QZ  =  PR  =  latitude  of  the  observer  =  <j>. 

The  triangle  ZPS  is  called  the  astronomical  triangle; 
ZP  =  90°  -  4»  =  co-latitude  of  the  observer, 


PS  =  90°  -  8,  SZ  =  90°  -  h. 


342  SPHERICAL   TRIGONOMETRY. 

Let  the  small  circle  MM',  passing  through  S,  and  parallel 
to  the  equinoctial,  represent  the  apparent  diurnal  motion  of 
the  heavenly  body  S  (the  declination  being  supposed  con- 
stant) ;  then  the  body  S  will  appear  to  rise  at  A  (if  we  sup- 
pose the  Eastern  hemisphere  is  represented  in  the  diagram). 
It  will  be  at  B  at  6  o'clock  in  the  morning,  at  M  at  noon, 

at  M'  at  midnight,  and  at  «  it  will  be  east. 

• 

225.  Problems.  —  By  means  of  the  foregoing  definitions 
and  diagram  we  may  solve  several  astronomical  problems  of 
an  elementary  character  as  follows : 

(1)  Given  the  latitude  of  a  place  and  the  declination  of  a 
star;  to  find  the  time  of  its  rising. 

Let  A  be  the  position  of  the  star  in  the  horizon.  Then 
in  the  triangle  APE,,  right  angled  at  E,  we  have 

cos  EPA  =  -  cos  ZPA  =  tan  EP  cot  AP. 

.*.  cos  t  =  —  tan  <£  tan  8 (1) 

from  which  the  hour  angle  is  found. 

Since  the  hourly  rate  at  which  a  heavenly  body  appears 
to  move  from  east  to  west  is  15°,  if  the  hour  angle  be 
divided  by  15  the  time  will  be  found.  In  the  case  of  the 
sun,  formula  (1)  gives  the  time  from  sunrise  to  noon,  and 
hence  the  length  of  the  day. 

Ex.  Eequired  the  apparent  time  of  sunrise  at  a  place 
whose  latitude  is  40°  36'  23".9,  on  July  4,  1881,  when  the 
sun's  declination  is  22°  52'  1". 


<£  =  40°  36'  23". 9, 

8  =  22°  52'    1". 


log  tan  4  =  9.9331352 
loff  tan  8  =  9.6250362 


log  cos  t  =  9.5581714- 
.-.   t  =111°  11' 44" 

=  7h  24m  47s,  nearly, 


PROBLEMS  OF  SPHERICAL  ASTRONOMY.       343 

which  taken  from  12h,  the  time  of  apparent  noon,  gives 
4h  35m  13s,  the  time  of  apparent  sunrise.* 

(2)  Given  the  latitude  of  a  place  and  the  declination  of  a 
star;  to  find  its  azimuth  from  the  north  at  rising. 

Let  A  =  the  azimuth  =  AR.  Then  in  the  triangle  APR 
we  have 

sin  AP  =  cos  AR  cos  PR, 

or  sin   8    =  cos  A  cos  <£. 

.*.  cos  A=  sin  0  sec  <£ (2) 

Ex.  Required  the  hour  angle  and  azimuth  of  Arcturus 
when  it  rises  to  an  observer  in  New  York,  lat.  40°  42'  N., 
the  declination  being  19°  57'  N. 

Ans.  7h  12m  468.3  ;  K  63°  15'  11"  E. 

(3)  Given  the  latitude  of  the  observer  and  the  hour  angle 
and  declination  of  a  star;  to  find  its  azimuth  and  altitude. 

Here  we  have  given,  in  the  triangle  ZPS,  two  sides  and 
the  included  angle;  that  is,  PZ  =  90°-<£,  PS  =  90°  —  8, 
and  ZPS  =  t.  Let  A  =  the  azimuth  from  the  north  =  RT, 
p  =  the  angle  ZSP,  and  z  =  ZS.  Then  by  Delambre's 
Analogies  (Art.  198), 

siu.%(p  +  A)  cos  % z  =  cos ^(8  —  </>)  cos^J, 
sin  ^(p  —  A)  sin  £z  =  sin  ^-(8  —  <£)cos^£, 
+  A)  cos  $z  =  sin  i(S+  <£)sin££, 
—  A)  sin  £z  =  cos |(8  +  <£)  sin^-Z. 

Hence,  when  <£,  t,  and  8  are  given,  that  is,  the  latitude  of 
the  place,  and  the  hour  angle  and  the  declination  of  a 
heavenly  body,  A,  z,  and  p  can  be  found. 

In  a  similar  manner  may  be  solved  the  converse  problem  : 
Given  the  latitude  of  the  observer  and  the  azimuth  and  altitude 
of  a  star;  to  find  its  hour  angle  and  declination. 

*  In  these  examples  no  corrections  are  applied  for  refraction,  serai-diameter  of 
the  sun,  change  in  declination  from  noon,  etc. 


344  SPHERICAL   TRIGONOMETRY. 

(4)  Given  the  right  ascensions  and  declinations  oftivo  stars; 
to  find  the  distance  between  them. 

Let  P  be  the  pole,  S  and  S'  the  two  stars.    p  a_a/ 
Let  a  and  a'  be  the  right  ascensions  of  the 
stars  ; 

8  and  8'  their  declinations  ;  and 

d  the  required  distance. 

Then  we  have  given,  in  the  triangle  PSS', 
two  sides  and  the   included   angle  ;   that  is,       s, 
PS  =  90°-S=p,      PS'  =  90°-S'=p',      and 
P  =  a  -  «'. 

This  may  be  solved  by  Art.  198,  or  by  the 
second  method  of  Art.  209,  as  follows  : 

Draw  SD  perpendicular  to  PS'  produced;  let  PD  =  m. 
Then 

cosP  =  tanPDcotPS. 

.-.  tan  m  =  cos  P  tan  p. 

Also  cos  SS'  =  cos  PS  cos  S'D  sec  PD    .    (Art.  209) 

.•.    cos  d  =  cos  p  cos  S'D  seem. 

Ex.  Required  the  distance  between  Sirius  and  Aldebaran, 
the  right  ascensions  being  6h  38m  378.6  and  4h  27™  258.9,  and 
the  declinations  16°  31'  2"  S.  and  16°  12'  27"  K,  respectively. 

Here  P  =  2h  llm  118.7  log  cos  P  =  9.9245789 

=    32°  47'  55"  log  tan  p  =  0.5279161- 

p  =  106°  31'    2"  log  tan  m  =  0.4524950- 

m=  109°  25'  55"  m  =  109°  25'  55". 

'=    73°47;33" 


S'D=    35°  38'  22" 

log  cos  S'D  =  9.9099302 

log  cos^>  =  9.4537823- 
m  =  109°  25'  55",  coi0g  cos  m  =  0.4779643- 

=  d=    46°    0'44".  log  cos  cZ  =  9.8416768 


=  106'31'    2", 


PROBLEMS  OF  SPHERICAL  ASTRONOMY.       345 


(5)   Given  the  right  ascension  and  declination  of  a  star;  to 
find  its  latitude  and  longitude. 

Let  V  be  the  vernal  equinox,  S  the  star,  VD,  VL  the 
the  equator  and  the  ecliptic,  SD,  SL 
perpendicular  to  VD,  VL.  Then 
VD  =  right  ascension  =  «,  SD  =  dec- 
lination =  8,  VL  =  longitude  =  A, 
SL  =  latitude  =  /.  Denote  the  ob- 
liquity of  the  ecliptic  DVL  by  <u, 
and  the  angle  DVS  by  0. 

From    the    right   triangles    SVD, 
SVL  we  get 

cot  6  =  sin  a  cot  8 (1) 

tan  A  =  cos  (0  —  to)  tan  a  sec  6      .....     (2) 
sin  I  =  sin  (6  —  w)  sin  8  cosec  0 (3) 

From  (1),  6  is  determined;  and  from  (2)  and  (3),  A  and 
I  are  determined. 

Ex.  Given  the  right  ascension  of  a  star  5h6m428.01,  and 
its  declination  45°  51'  20".l  N.;  to  find  its  longitude  and 
latitude,  the  obliquity  of  the  ecliptic  being  23°  27'  19".45. 


a  =  76°  40'  30".15 
8  =  45°  51'  20".l 
0  =  46°38'11".8 
a)  =  23°  27'  19".45 

0  -  a)  =  23°  10'  52".35 


A  =  79°  58'    3".44. 


J=22°51'48".4 


log  sin  a  =    9.9881479 
log  cot  8  =    9.9870277 

log  cot  9  =    9.9751756 

log  cos  (0  -  a>)  =    9.9634401 

log  tan  a  =  10.6255266 

cologcos<9=    0.1632816 

log  tan  A  =10.7522483  s 

log  sin  (6  -  «)  =    9.5950996 

log  sin  8  =    9.8558743 

cologsin<9=    0.1384575 

log  sin/  =    9.5894314 


346  SPHERICAL    TRIGONOMETRY. 


EXAMPLES. 

1.  Find  the  apparent  time  of  sunrise  at  a  place  whose 
latitude  is  40°  42',  when  the  sun's  declination  is  17°  49'  N. 

Ans.  4h  56m. 

2.  Given  the  latitude  of  a  place  =  40°  36'  23".9,  the  hour 
angle  of  a  star=46°40'4".5,  and  its  declination  =  23°  4'  24".3 ; 
to  find  its  azimuth  and  altitude. 

Ans.  Azimuth  =  80°  23'  4  ".47,  altitude  =  47°  15'  18".3. 

3.  Find  the  altitude  and  azimuth  of  a  star  to  an  observer 
in  latitude  38°  53'  1ST.,  when  the  hour  angle  of  the  star  is 
3h  15m  20s  W.,  and  the  declination  is  12°  42'  K 

Ans.  Altitude  =  39°  38'  0";  azimuth  =  S.  72°  28'  14"  W. 

4.  Given  the  latitudes  of  New  York  City  and  Liverpool 
40°  42'  44"  N.  and  53°  25'  K,  respectively,  and  their  longi- 
tudes 74°0'24"W.    and  3°  W.,  respectively;   to  find   the 
shortest  distance  on  the  earth's  surface  between  them  in 
miles,   considering  the  earth   as   a  perfect   sphere  whose 
radius  is  3956  miles. 

NOTE.  —  This  is  evidently  a  case  of  (4)  where  two  sides  and  the  included  angle 
are  given,  to  find  the  third  side. 

Ans.  3305  miles. 

5.  The  latitudes  of  Paris  and  Pekin  are  48°  50'  14"  N. 
and   39°  54'  13"  K,  and   their   difference   of   longitude   is 
114°  7'  30";  find  the  distance  between  them  in  degrees. 

'Ans.  73°  56' 40". 

GEODESY. 

226.  The  Chordal  Triangle.  —  Given  tivo  sides  and  the 
included  angle  of  a  spherical  triangle;  to  find  the  correspond- 
ing angle  of  the  chordal  triangle. 


GEODESY. 


347 


The  chordal  triangle  is  the  triangle  formed 
by  the  chords  of  the  sides  of  a  spherical 
triangle. 

Let  ABC  be  a  spherical  triangle,  0  the 
centre  of  the  sphere,  A'BC  the  colunar 
triangle,  and  M,  N  the  middle  points  of 
the  arcs  A'B,  A'C.  Then  the  chord  AB  is 
parallel  to  the  radius  OM,  since  they  are 
both  perpendicular  to  the  chord  A'B.  Simi- 
larly, AC  is  parallel  to  ON". 

In  the  spherical  triangle  A'MN",  we  have 

cos  MN  =  cos  A'N  cos  A'M  +  sin  A'X  sin  A'M  cos  A'  (Art.  191) 

Denote  the  angle  BAG  of  the  chordal  triangle  by  Aj. 
Then    arc   WN   or   angle   MON  =  A1?    AfN  =  !(«•-&), 
A'M  =  J(ir-c),  and  A'  =  A. 


.-.  cos  A!  =  sin  -J-  b  sin  £  c-f-  cos  %  b  cos  -J-c  cos  A 
with  similar  values  for  cos  B!  and  cos  C^ 


(1) 


Cor.  1.    If  the   sides  b  and  c  are  small  compared  with 
the  radius  of  the  sphere,  Al  will  not  differ  much  from  A. 

Let  Aj  =  A  —  0;  then 

cos  AJ  =  cos  A  -f-  0  sin  A,  nearly. 
But      sin-|-&  sin-J-c  =  sin2  J(6  +  c)  —  sin2  1(6  —  c), 
and          cos  ^6  cos  £c  =  cos2  \(b  +  c)  —  sin2  J 
Substituting  in  (1)  and  reducing,  we  get 


which  is  the  circular  measure  of  the  excess  of  an  angle  of  the 
spherical  triangle  over  the  corresponding  angle  of  the  chordal 
triangle. 

The  value  in  seconds  is  obtained  by  dividing  the  circular 
measure  by  the  circular  measure  of  one  second,  or,  approxi- 
mately, by  the  sine  of  one  second. 


348  SPHERICAL    TRIGONOMETRY. 

Cor.  2.  The  angles  of  the  ch  or  dal  triangle  are,  respectively, 
equal  to  the  arcs  joining  the  middle  points  of  the  sides  of  the 
colunar  triangles. 

227.  Legendre's  Theorem.  —  If  the  sides  of  a  spherical 
triangle  be  small  compared  ivith  the  radius  of  the  sphere,  then 
each  angle  of  the  spherical  triangle  exceeds  by  one-third  of  the 
spherical  excess  the  corresponding  angle  of  the  plane  triangle, 
the  sides  of  which  are  of  the  same  length  as  the  arcs  of  the 
spherical  triangle. 

Let  a,  b,  c  be  the  lengths  of  the  sides  of  the  spherical 
triangle,  and  r  the  radius  of  the  sphere ;  then  the  circular 

measures  *  of  the   sides  are  respectively  -,  -,  —      Hence, 

r  r  r 

neglecting  powers  of  —  above  the  fourth, 

a  be 

cos cos  -  cos  - 

cosA  = r  T  c— (Art.  191) 

sin -sin - 
r       r 

te  /  ^  J*r/    !\      —  (Art- 156> 

72 


bc\        2r*  24  r4  A  Or2 


q__c_c 


26c  246CT8  6?'2 


ca 


2  be 

*  The  term  ?  is  the  circular  measure  of  the  angle  which  the  arc  a  subtends  at  the 
centre  of  the  sphere;  and  similarly  for  _  and  £. 


GEODESY.  349 

Now  if  A',  B'?  C'  denote  the  angles  of  the  plane  triangle 
whose  sides  are  a,  b,  c,  respectively,  we  have 


2  be 


(Art.  96) 


A,  -----       , 

and    sin2  A'  =  —  -  (Art.  100) 

4crc~ 

Therefore  (1)  becomes 

coSA  =  coSA'-6c^A'     ....     (2) 

Let  A  =»  A'  +  0,  where  0  is  a  very  small  quantity  ;  then 
cos  A  —  cos  A'  —  0  sin  A',  nearly. 


where  A  denotes  the  area  of  the  plane  triangle  whose  sides 

are  a,  6,  c. 

^ 
We  have  therefore  A  =  A'  -\  --  -• 

Similarly      B  =  B'  +  A  C  =  C'  +  A 
3r2  Sr2 

...  A  +  B  +  C  -  A'  -  B'  -  C'  =  -; 

r2 

or  A  +  B  +  C  —  TT  =  ~    =  spherical  excess  (Art.  219) 

.-.  A-A'-B-B'=C-C'-:^2  =  i  spherical  excess. 

3  1 

Cor.  1.  If  the  sides  of  a  spherical  triangle  be  very  small 
compared  with  the  radius  of  the  sphere,  the  area  of  the 
spherical  triangle  is  approximately  equal  to 


350  SPHERICAL    TRIGONOMETRY. 


For,  tanJE=A/tan  —  tan^Uan— tan  —  (Art.  220) 
\       2r         2r  2r          2r 

and 

far,     S    __    S    I   -1  _i        °        I  .    for,  "  — *-"  _  1~~"'|   1    i     V"~~"VV     I     PrP 

lic*'11  ~         ~        I     -L  ~T~     .  „       «     I  »      tctll       —  -       _  I     J-  T™  t)          U     CUO. 

(Art.  156) 


(Arts.  101  and  156) 


A/t  -L  r6-  +  &2  +  c2V_  A       _L 
~  ~( 


12  r2      y      4r2i  24?- 


That  is  :  the  area  of  the  spherical  triangle  exceeds  the  area 
the  plane  triangle  by  -  "*"     /"  C  parf  of  the  latter. 

Cor.  2.  If  we  omit  terms  of  the  second  degree  in  -,  we  have 


Hence,  if  the  sides  of  a  spherical  triangle  be  very  small 
compared  with  the  radius  of  the  sphere,  its  area  is  approxi- 
mately equal  to  the  area  of  the  plane  triangle  having  sides 
of  the  same  length. 

228.  Roy's  Rule.  —  The  area  of  a  spherical  triangle  on  the 
Earth's  surface  being  known,  to  establish  a  formula  for  com- 
puting the  spherical  excess  in  seconds. 

Let  A  be  the  area  of  the  triangle  in  square  feet,  and  n  the 
number  of  seconds  in  the  spherical  excess.  Then  we  have 


180°  x  60  x  60 

nr2 


ROY'S  RULE.  351 

.     .     .  (Art.  219) 

_  (1} 

180  x  60  x  60     206265  " 

Now,  the  length  of  a  degree  on  the  Earth's  surface  is 
found  by  actual  measurement  to  be  365155  feet. 

...   -^=365155.    ..  r==  180x365155. 

180°  .         7T 

Substituting  this  value  of  r  in  (1),  and  reducing,  we  get 
log  n  =  log  A  -9.3267737 (2) 

This  formula  is  called  General  Roy's  rule,  as  it  was  used 
by  him  in  the  Trigonometric  Survey  of  the  British  Isles. 
He  gave  it  in  the  following  form :  From  the  logarithm  of 
the  area  of  the  triangle,  taken  as  a  plane  triangle,  in  square 
feet,  subtract  the  'constant  logarithm  9.3267737 ;  and  the  re- 
mainder is  the  logarithm  of  the  excess  above  180°,  in  seconds, 
nearly. 

Ex.  If  the  observed  angles  of  a  spherical  triangle  are 
42°  2' 32",  67°  55' 39",  70°1'48",  and  the  side  opposite  the 
angle  A  is  27404.2  feet,  required  the  number  of  seconds  in 
the  sum  of  the  errors  made  in  observing  the  three  angles. 

Here  the  apparent  spherical  excess  is 

A  +  B  +  C  -  180°  =  -  1". 

The  area  of  the  triangle  is  calculated  from  the  expression 

a2sinBsinC  (Art.  101) 

2  sin  A 

and  by  Roy's  Rule  the  computed  spherical  excess  is  found 
to  be  .23". 

Now  since  the  computed  spherical  excess  may  be  supposed 
to  be  the  real  spherical  excess,  the  sum  of  the  observed 
angles  ought  to  have  been  180°  -f  .23". 

Hence  it  appears  that  the  sum  of  the  errors  of  the  obser- 
vations is  .23"  — (-!")  =  1".23,  which  the  observer  must 


352  SPHERICAL    TRIGONOMETRY. 

add  to  the  three  observed  angles,  in  such  proportions  as  his 
judgment  may  direct.  One  way  is  to  increase  each  of  the 
observed  angles  by  one-third  of  1".23,  and  take  the  angles 
thus  corrected  for  the  true  angles. 

229.  Reduction  of  an  Angle  to  the  Horizon.  —  Given  the 
angles  of  elevation  or  depression  of  two  objects,  which  are  at  a 
small  angular  distance  from  the  horizon,  and  the  angle  which 
the  objects  subtend,  to  fyid  the  horizontal  angle  between  them. 

Let  a,  b  be  the  two  objects,  the  angular  distance  between 
which  is  measured  by  an  observer  at  ^ 

O ;  let  OZ  be  the  direction  at  right 
angles  to  the  observer's  horizon.  De- 
scribe a  sphere  round  O  as  a  centre, 
and  let  vertical  planes  through  Oa, 
06,  meet  the  horizon  at  OA,  OB,  re- 
spectively ;  then  the  horizontal  angle 
AOB,  or  AB,  is  required. 

Let    ab  =  0,    AB  =  6  -f  x,    Aa  =  h, 
B6  =  It.     Then  in  the  triangle  aZb  we  have 

cos  ab  —  cos  aZ  cos  bZ 


cos  AB  =  cos  aZb  = 
or  cos  (6  H-  x)  = 


sin  aZ  sin  6Z 

cos  0  —  sin  h  sin  k 
cos  h  cos  k 


This  gives  the  exact  value  of  AB  ;  by  approximation  we 
obtain,  where  x  is  essentially  small, 

.    n          cos  6  —  hk 


.-.  x  sin  0  =  hk  —  £  (h2  +  A;2)  cos  0,  nearly. 
2  hk  -  (h2  +  k2)  /"cos2  1  -  sin2 
"'•  X=  2sin0 

=  i  CO  +  &)8  tan  i  0  -  (ft  -  &)2  cot  i  0]. 


SMALL    VARIATIONS  IN  PARTS   OF  TRIANGLES.      353 

EXAMPLES. 

1.  Prove  that  the  angles  subtended  by  the  sides  of    a 
spherical  triangle  at  the  pole  of  its  circumcircle  are  respec- 
tively double  the  corresponding  angles  of  its  chordal  tri- 
angle. 

2.  If  AU  B1?  Ci ;  A2,  B2,  C2 ;  A3,  B3,  C3 ;  be  the  angles  of 
the  chordal  triangles  of  the  colunars,  prove  that 

cos  A^cos^asinS,  cos  B^  sin  £  &sin(S  —  C),  cos  C^  sin  2  csin(S  — B), 

cosA2=:sin2  asin(S  — C),  cos  B2  =  cos -I  &sinS,  cosC2  =  sin2Csin(S  —  A), 

j^sin^  asin(S— B),  cosB3=sin.l  6sin(S  — A),  cosC3=c 


3.  Prove  Legendre's  Theorem  from  either  of  the  formulae 
for  sin  |-  A,  cos  \  A,  tan  J  A,  respectively,  in  terms  of  the 
sides. 

4.  If  C  =  A  +  B,  prove  cos  C  =  —  tan  £  a  tan  -J-  b. 

230.  Small  Variations  in  the  Parts  of  a  Spherical  Tri- 
angle. 

It  is  sometimes  important  in  Geodesy  and  Astronomy  to 
determine  the  error  introduced  into  one  of  the  computed 
parts  of  a  triangle  from  any  small  error  in  the  given  parts. 

If  two  parts  of  a  spherical  triangle  remain  constant,  to  deter- 
mine the  relation  between  the  small  variations  of  any  other  tivo 
parts. 

Suppose  C  and  c  to  remain  constant. 

(1)  Eequired  the  relation  between  the  small  variations 
of  a  side  and  the  opposite  angle  (a,  A). 

Take  the  equation 

sin  A  sine  =  sin  C  sin  a (1) 

We  suppose  a  and  A  to  receive  very  small  increments  da 
and  c?A;  then  we  require  the  ratio  of  da  and  dA  when  both 
are  extremely  small.  Thus 

H^  sin  (A  +  dA)sinc  =  sinC  sin  (a 


354  SPHERICAL   TRIGONOMETRY. 

or  (sin  A  cos  dA  -f  cos  A  sin  dA)  sin  c 

=  sin  C  (sin  a  cos  da  +  cos  a  sin  da) (2) 

Because  the  arcs  dA  and  da  are  extremely  small,  their 
sines  are  equal  to  the  arcs  themselves  and  their  cosines 
equal  1 :  therefore  (2)  may  be  written 

sin  A  sin  c  +  cos  A  sin  cdA  =  sin  C  sin  a  -f  sin  C  cos  ada     (3) 
Subtracting  (1)  from  (3),  we  have 
cos  A  sin  cdA  =  sin  C  cos  ada. 

da  _  cos  A  sin  c  _  tan  a 
dA      sin  C  cos  a      tan  A 

(2)  Required  the  relation  between  the  small  variations 
of  the  other  sides  (a,  b).     We  have 

cos  c  =  c«s  a  cos  b  -f  sin  a  sin  b  cos  C (1) 

.•.  cosc  =  cos(a-j-da)cos(6-f-  db)  -|-siii(a-|-da)sin(&-4-d6)cosC, 

or  =  (cos  a  —  sin  ada)  (cos  b  —  sin  bdb) 

4-  (sin  a  +  cos  ada)  (sin  6  -f  cos  bdb)  cos  C  .     (2) 

Subtracting  (2)  from  (1)  and  neglecting  the  product 
da  db,  we  have 

0  =  (sin  a  cos  b  —  cos  a  sin  6  cos  C)  da 

+  (cos  a  sin  fr  —  sin  a  cos  6  cos  C)  a7>, 

.v  _(cot6sina— cosacosC)  ,       (cot  a  sin  6— cos  6  cos  C)  „ 
sin  a  sin  6 

^  _  cot  B  sin  C  da      cot  A  sin  C  db  (\t  1931 

sin  a  sin  6 

da  _      cos  A 
db          cos  B 

(3)  Required  the  relation  between  the  small  variations 
of  the  other  angles  (A,  B). 


SMALL  VARIATIONS  IN  PARTS  OF  TRIANGLES.      355 

By  means  of  the  polar  triangle,  we  may  deduce  from  the 
result  just  found,  that 

dA  _      cos  a 
dB          cos  b 

(4)  Required  the  relation  between  the  small  variations 
of  a  side  and  the  adjacent  angle  (b,  A).     We  have 

cot  c  sin  b  =  cot  C  sin  A  +  cos  b  cos  A  ...     (Art.  193) 

Giving  to  b  and  A  very  small  increments,  and  subtracting, 
as  before,  we  get 

cot  c  cos  bdb  =  cot  C  cos  AdA  —  sin  b  cos  Adb  —  cos  b  sin  Ad  A. 
(cote  cos  b  +  sin  6  cos  A)db  =  (cot  C  cos  A  —  cos  b  sin  A)  dA. 

db  =  _  ^^  da  .     .     (Arts.  191  and  192) 


sin  c  sin  C 

db  _  _  cos  B  sin  b  _      sin  b  cot  B 
dA          cos  a  sin  B  cos  a 


EXAMPLES. 

1.  If  A  and  c  are  constant,  prove  the  following  relations 
between  the  small  variations  of  any  two  parts  of  the  other 
elements : 

da  _       tan  a  t  db  _  sin  a 

dC  ~  ~  tanC  '  dB  ~~  sinC* 

db  _  _  tan  a  m  dC  _  _ 

dC  ~  ~  sin  C  '  dB~ 

2.  If  B  and  C  remain  constant,  prove  the  following : 

«=ten6  *±  =  sin  B  sinC. 

dc       tan  c  da 

*»  sina 


dc  dc       sin  c  cos  6 


156 


SPHERICAL    TRIG  ONOMETR  Y. 


POLYEDRONS. 

231.   To  find  the  Inclination  of  Two  Adjacent  Faces  of  a 
Regular  Polyedron. 

Let  C  and  D  be  the  centres  of  the  cir- 
cles inscribed  in  the  two  adjacent  faces 
whose  common  edge  is  AB ;  bisect  AB 
in  E,  and  join  CE  and  DE ;  CE  and  DE 
will  be  perpendicular  to  AB.  .-.  Z  CED 
is  the  inclination  of  the  two  adjacent 
faces,  which  denote  by  I. 

In  the  plane  CED  draw  CO  and  DO  at 
right  angles  to  CE  and  DE,  respectively, 
and  meeting  in  0.  Join  OA,  OE,  OB,  and  from  0  as  centre 
describe  a  sphere,  cutting  OA,  OC,  OE  at  a,  c,  e,  respec- 
tively ;  then  ace  is  a  spherical  triangle.  Since  AB  is  per- 
pendicular to  CE  and  DE,  it  is  perpendicular  to  the  plane 
CED;  therefore  the  plane  AOB,  in  which  AB  lies,  is  per- 
pendicular to  the  plane  CED.  .-.  Z  aec  is  a  right  angle. 

Let  m  be  the  number  of  sides  in  each  face,  and  n  the 
number  of  plane  angles  in  each  solid  angle.     Then 


- 
2m 


m 


and 


Zcae  =  |Z  of  the  planes  OAC  and  OAD. 


In  the  right  triangle  cae  we  have 

cos  cae  —  cos  ce  sin  ace. 
But 


cos  ce  =  cos  COE  =  cos  ( ]=  sin--- 


2     2 


cos- =  sin -sin— 


n 


m 


sin  -  =  cos  -  cosec  — • 
n          m 


VOLUME  OF  A  PARALLELOPIPED.  357 

Cor.  1.    If  r  be  the  radius  of  the  inscribed  sphere,  and  a  be 
a  side  of  one  of  the  faces,  then 

r  = -cot-tan-. 
2       m        2 

For,  r  =  OC  =  CE  tan  CEO  =  AE  cot  ACE  tan  CEO 

=  2COtmtan2' 
Cor.  2.    J/*R  be  the  circumradius  of  the  polyedron,  then 

R  =  -  tan-  tan-- 

2        n        2 

For,  r  =  0 A  cos  aoc  =  R  cot  eca  cot  eac  =  R  cot—  cot  -• 

m       n 

.-.  R  =  -tan^tan-. 
2        n       2 

Cor.  3.    The  surface  of  a  regular  polyedron,  F  being  the 

number  of  faces,    =t  —    —  cot—  • 
4          m 

For,  the  area  of  one  face  =  —  m  cot— «     .•.  etc. 

4  m 

Cor.  4.    TVie  volume  of  a  regular  polyedron 


12          m 

For,  the  volume  of  the  pyramid  which  has  one  face  of 
the  polyedron  for  base  and  0  for  vertex 


r  ma2     ,  TT 

cot  —  -     .-.  etc. 

34         m 


232.  Volume  of  a  Parallelepiped.  —  To  find  the  volume  of 
a  parallelepiped  in  terms  of  its  edges  and  their  inclinations 
to  one  another. 


358  SPHERICAL   TRIGONOMETRY. 

Let  the  edges  be  OA  =  a, 
OB  =  b,  OC  =  c,  and  let  the 
inclinations  be  BOC  =  a,  COA 
=  ft  AOB  =  y.  Draw  CH 
perpendicular  to  the  face 
AOBE.  Describe  a  sphere  o< 
round  0  as  centre,  meeting 
OA,  OB,  OC,  OE,  in  a,  b,  c,  e, 
respectively. 

The  volume  of  the  parallelo-  E 

piped  is  equal  to  the  area  of  the  base  OAEB  multiplied  by 
the  altitude  CH ;  that  is, 

volume  =  ab  sin  y  •  CH  =  dbc  sin  y  sin  ce 
where  ce  is  the  perpendicular  arc  from  c  on  ab. 

.'.  volume  =  dbc  sin  y  sin  ac  sin  bac     .     .     .     (Art.  186) 

9m 

=  abc  sin  y  sin  ft — (Art.  195) 

H  sin  ft  sin  y 

=  abc  Vl  —  cos2«  —  cos2 (3  —  cos2  y  -f  2  cos  a  cos  ft  cos  y. 
(7or.  1.    The  surface  of  a  parallelepiped 

=  2  (6c  sin  «  -f-  ca  sin  ft  -j-  aft  sin  y). 
Cor.  2.    T/ie  volume  of  a  tetraedron 

=  i  abc  Vl  —  cos2  a  —  cos2  /? — cos2  y + 2  cos  a  cos  /?  cos  y. 

For,  a  tetraedron  is  one-sixth  of  a  parallelepiped  which 
has  the  same  altitude  and  its  base  double  that  of  the 
tetraedron. 

233.  Diagonal  of  a  Parallelepiped. —  To  find  the  diagonal 
of  a  parallelepiped  in  terms  of  its  edges,  and  their  mutual 
inclinations. 

Let  OD  (figure  of  Art.  232)  be  a  parallelepiped,  whose 
edges  OA  =  a,  OB  =  6,  OC  =  c,  and  their  inclinations  BOC 
=  «,  COA  =  ft,  AOB  =  y ;  let  OD  be  the  diagonal  required, 


TABLE   OF  FORMULA.  359 

and  OE  the  diagonal  of  the  face  OAB.     Then  the  triangle 
OED  gives 

OD2  =  OE2  +  ED2  +  2  OE  -  ED  cos  COE 

=  a2  +  tf  +  2  ab  cos  y  +  c2  +  2c  •  OE  cos  COE       (1) 

Now,  it  is  clear  that  OE  cos  COE  is  the  projection  of  OE 
on  the  line  OC,  and  therefore  it  must  be  equal  to  the  sum 
of  the  projections  of  OB  and  BE  (or  of  OB  and  OA),  on 
the  same  line.* 

.-.  OE  cos  COE  =  b  cos  a  +a  cos  /?, 
which  in  (1)  gives 

OD2  =  a2  +  b-  +  c2  -f  2bc  cos  «  +  2ca  cos  p  +  2ab  cos  y  .     (2) 


234.   Table  of  Formulae  in  Spherical  Trigonometry.  —  For 

the    convenience   of   the    student,    many   of   the   preceding 
formulae  are  summed  up  in  the  following  table  : 

1.  cos  c  =  cos  a  cos  b    ........     (Art.  185) 

2.  sin  b  =  sin  B  sin  c. 

3.  sin  a  =  sin  A  sine. 

4.  cos  C  =  —  cos  A  cos  B  .......     (Art.  189) 

5.  sin  B  =  sin  6  sin  C. 

6.  sin  A  =  sin  a  sin  C. 

r_   1in^=shL6=iinc  (Art.  190) 

sin  A      sin  B      sin  C 

8.  cos  a  =  «os  b  cose  -f  sin  b  sine  cos  A  .     .     (Art.  191) 

9.  cos  b  =  cos  c  cos  a  +  sin  c  sin  a  cos  B. 

10.  cos  c  =  cos  a  cos  b  +  sin  a  sin  6  cos  C. 

11.  cos  A  =  —cos  B  cos  C  +  sin  B  sin  C  cos  a       (Art.  192) 

12.  cos  B  =  —  cos  C  cos  A  -f  sin  C  sin  A  cos  b. 

*  From  the  nature  of  projections  (Plane  and  Solid  Geom.,  Art.  326). 


360  SPHERICAL    TRIGONOMETRY. 

13.  cos  C  =  —  cos  A  cos  B  -f  sin  A  sin  B  cos  c. 

14.  cot  a  sin  b  =  cot  A  sin  C  -f  cos  C  cos  b  .     .     (Art.  193) 

15.  cot  a  sin  c  =  cot  A  sin  B  -f  cos  B  cos  c. 

16.  cot  b  sin  a  =cot  B  sin  C  -f  cos  C  cos  a. 

17.  cot  b  sin  c  =  cot  B  sin  A  -f  cos  A  cos  c. 

18.  cot  c  sin  a  =  cot  C  sin  B  -f-  cos  B  cos  a. 

19.  cot  c  sin  b  =  cot  C  sin  A  -f  cos  A  cos  b. 

20.  sin  a  cos  B  =  cos  b  sin  c  —  sin  b  cos  c  cos  A    (Art.  194) 

21.  sin  a  cos  C  =  sin  b  cos  c  —  cos  6  sin  c  cos  A. 

22.  sin  b  cos  A  =  cos  a  sin  c  —  sin  a  cos  c  cos  B. 

23.  sin  b  cos  C  =  sin  a  cos  c  —  cos  a  sin  c  cos  B. 

24.  sin  c  cos  A  =  cos  a  sin  b  —  sin  a  cos  b  cos  C. 

25.  sin  c  cos  B  =  sin  a  cos  6  —  cos  a  sin  b  cos  C. 


26.   sin^A=Jsi"(s-6)sin(s-c)     .     .     .     (Art.  195) 
\  sin  b  sin  c 


sin  b  sin  c 
27. 


sin  b  sin  c 
28.   tan  1 A 


sin  s  sin  (s  —  a) 


29    sin  A 


~  c) 


sin  6  sin  c 


sin  6  sin  c 


where  n  =  Vsin  s  sin  (s  —  a)  sin  (s  —  b)  sin  (s  —  c). 

30.   sin4a=J-cosScos(S-AJ  (Art.  196) 

\          sin  B  sin  O 


31.   cos i a  =  JcoLi8_BlcoB{S_CJ 
\  sin  B  sin  O 


TABLE  OF  FORMULAE.  361 


00  cos  (S  —  A) 

32.  - 


-C) 


QQ      .  _2V  —  cosScos(S  —  A)  cos(S  —  B)cos(S  —  C) 

oo.     Sill  Cf       —  —  —  ;      ~ 

sin  B  sin  C 

34.  tan  KA  +  B)  =  cos  Kft  -  &)  cot  ^Q    .     .     (Art.  197) 

cos  |(a  +  6) 

35.  ta 


sin 


37-  tan*(a-6)  = 

38.  sin£(A  +  B)cos!c  =  cosi(a-&)cosiC   (Art.  198) 

39.  sin  J(A  -  B)  sin  Jc  =  sin  ^(a  -  6)  cos  JC. 

40.  cos  |(  A  +  B)  cos  J  c  =  cos  £(a  +  6)  sin  £  C. 

41.  cos  J(A  -  B)  sin  £c  =  sin  J(a  +  &)  sin  JC. 


=  v  p 
\ 


42.    tanr  ~  a   sin   g  ~  6   sn 


sns 


sins 
......     (Art.  215) 


43.  tanK  =  —  [sin(s—  a)  +sin(s—  b)  -fsin(s—  c)  —sins] 

(Art.  217) 

44.  K  =  areaof  A  =  -^l-Trr2    ......     (Art,  219) 

180 

45.  sin  |  E  =  -  ...     (Art.  220) 

2  cos  %  a  cos  f  o  cos  £  c 

46.  tan  \  E  =  Vtan£stan£(s—  a)  tan£  (s  —  b)  tan  J(s—  c). 


362  SPHERICAL   TRIGONOMETRY. 

EXAMPLES. 

1.  Find  the  time  of  sunrise  at  a  place  whose  latitude  is 
42°  33'  N.,  when  the  sun's  declination  is  13°  28'  N. 

Ans.  5h9m13V 

2.  Find  the  time  of  sunset  at  Cincinnati,  lat.  39°  6'  1ST., 
when  the  sun's  declination  is  15°  56'  S.  Ans.  5h  6in. 

3.  Find  the  time  of  sunrise  at  lat.  40°  43'  48"  N.,  in  the 
longest  day  in  the  year,  the  sun's  greatest  declination  being 
23°  27'  N.  Ans.  4h  32m  168.4. 

4.  Find  the  time  of  sunrise  at  Boston,  lat.  42°  21' N., 
when  the  sun's  declination  is  8°  47'  S.  Ans.  6h  14m. 

5.  Find  the  length  of  the  longest  day  at  lat.  42°  16'  48".3 
N.,  the  sun's  greatest  declination  being  23°  27'  N. 

Ans.  15h5m501. 

6.  Find  the  length  of  the  shortest  day  at  New  Bruns- 
wick, N.  J.,  lat.  40°  29'  52". 7  N.,  the  sun's  greatest  declina- 
tion being  23°  27'  S. 

7.  Find  the  hour  angle  and  azimuth  of  Antares,  declina- 
tion 26°  6'  S.,  when  it  sets  to  an  observer  at  Philadelphia, 
lat.  39°  57'  N.  Ans.  4h  23m  59.7 ;  S.  54°  58'  44"  W. 

8.  Find  the  hour  angle  and  azimuth  of  the  Nebula  of 
Andromeda,  declination  40°  35'  N.,  when  it  rises  to  an  ob- 
server at  New  Brunswick,  N.  J.,  lat.  40°  29'  52".7  N. 

9.  Find  the  azimuth  and  altitude  of  Regulus,  declination 
16°  13'  N.,  to  an  observer  at  New  York,  lat.  40°  42'  N.,  when 
the  star  is  three  hours  east  of  the  meridian. 

Ans.  Azimuth  =  S.  71°  12'  30"  E. ;  Altitude  =  44°  10'  33". 

10.  Find  the  azimuth  and  altitude  of  Fomalhaut,  dec- 
lination 30°  25'  S.,  to  an  observer  in  lat.  42°  22'  N.,  when  the 
star  is  2h  5m  36s  east  of  the  meridian. 

Ans.  Azimuth  =  S.  27°  18'  40"  E. ;  Altitude  =  11°  41'  37". 


EXAMPLES.  363 

11.  Find  the  azimuth  and  altitude  of  a  star  to  an  observer 
in    lat.    39°  57'  N.,    when    the    hour   angle    of    the    star   is 
5h  17m  40s  east,  and  the  declination  is  62°  33'  N. 

Ans.  Azimuth  =  N.  35°  54'  E. ;  Altitude  =  39°  24'. 

12.  Find  the  hour  angle  (t)  and  declination  (8)  -of  a  star 
to  an  observer  in  lat.  40°  36'  23".9  N.,  when  the  azimuth  of 
the  star  is  80°  23'  4".47,  and  the  altitude  is  47°  15'  18".3. 

Ans.  t  =  46°  40'  4".53 ;  8  =  23°  4'  24".33. 

13.  Find  the  distance  between  Regulus  and  Antares,  the 
right  ascensions  being  10h  Om  29M1  and  16h  20m  208.35,  and 
the  polar  distances  77°  18'  41".4  and  116°  5'  55".5. 

Ans.  99°55'44".9. 

14.  Find  the  distance  between  the  sun  and  inoon  when 
the  right  ascensions  are  12h  39m  38.22  and  6h  55m  32s. 73,  and 
the  declinations  9°  23'  16".7  S.  and  22°  50'  21".9  K 

Ans.  89°52'55".5. 

15.  Find  the  shortest  distance  on  the  earth's  surface,  in 
miles,  from  New  York,  lat.  40°  42'  44"  K,  long.  74°  0'  24"  W., 
to  San  Francisco,  lat.  37°  48'  N.,  long.  122°  23'  W. 

Ans.  2562  miles. 

16.  Find  the  shortest  distance  on  the   earth's   surface 
from   San   Francisco,  lat.   37°  48'  N.,  long.   122°  23' W.,  to 
Port  Jackson,  lat.  33°  51'  S.,  long.  151°  19'  E. 

Ans.  6444  nautical  miles. 

17.  Given  the  right  ascension  of  a  star  10h  lm  9S.34,  and 
its  declination  12°  37'  36".8  K  ;  to  find  its  latitude  and  longi- 
tude, the  obliquity  of  the  ecliptic  being  23°  27'  19".45. 

Ans.  Latitude  =  ;  Longitude  = 

18.  Given  the  obliquity  of  the  ecliptic  w,  and  the  sun's 
longitude  A ;  to  find  his  right  ascension  a  and  declination  8. 

Ans.  tan  a  —  cos  <o  tan  \ ;  sin  8  =  sin  o>  sin  A. 


364  SPHERICAL    TRIGONOMETRY. 

19.  Given  the  obliquity  of  the  ecliptic  23°  27'  18".5,  and 
the  sun's  longitude  59°  40f  1".6;  to  find  his  right  ascension 
(a),  and  declination  (8). 

Ans.  a  =  3h  49m  528.62 ;  8  =  20°  5' 33".9  K 

20.  Given  the  sun's  declination  16°  0'  56".  4  K,  and  the 
obliquity  of  the  ecliptic  23°27'18".2;   to   find   his   right 
ascension  («),  and  longitude  (A). 

Ans.  a  =  9h  14m  198.2  ;  A  =  136°  7'  6".5. 

21.  Given  the  sun's  right  ascension  14h  8m  19S.06,  and  the 
obliquity  of  the  ecliptic  23°  27'  17".8 ;  to  find  his  longitude 
(A),  and  declination  (8). 

Ans.  X  =  214°  20'  34".  7 ;  8  =  12°  58'  34".4  S. 

22.  Given    the   sun's   longitude   280°  23'  52". 3,    and   his 
declination  23°  2'  52".2  S. ;  to  find  his  right  ascension  (a). 

Ans.  a  =  18h  45m  14s. 7. 

23.  In  latitude  45°  N.,  prove  that  the  shadow  at  noon  of 
a  vertical  object  is  three  times  as  long  when  the  sun's  dec- 
lination is  15°  S.  as  when  it  is  15°  N. 

24.  Given  the  azimuth  of  the  sun  at  setting,  and  also  at 
6  o'clock ;  find  the  sun's  declination,  and  the  latitude. 

25.  If  the  sun's  declination  be  15°  N.,  and  length  of  day 
four  hours,  prove  tan  <£  =  sin  60°  tan  75°. 

26.  Given  the  sun's  declination  and  the  latitude ;  show 
how  to  find  the  time- when  he  is  due  east. 

27.  If  the  sun  rise  northeast  in  latitude  <£,  prove  that  cot 
hour  angle  at  sunrise  =  —  sin  <£. 

28.  Given  the  latitudes  and  longitudes  of  two  places ; 
find  the  sun's  declination  when  he  is  on  the  horizon  of  both 
at  the  same  instant. 

29.  Given    the    sun's    declination    8,    his   altitude   h   at 
6    o'clock,    and    his    altitude    h'   when    due    east;    prove 
sin2  8  =  sin  h  sin  h'. 


EXAMPLES.  365 

30.  Given  the  declination  of   a  star  30°  ;   find  at  what 
latitude  its  azimuth  is  45°  at  the  time  of  rising. 

31.  Given  the  sun's  declination  8,  and  the  latitude  of  the 
place  <£  ;  find  his  altitude  when  due  east. 

32.  Given  the  declinations  of  two  stars,  and  the  differ- 
ence of  their  altitudes  when  they  are  on  the  prime  vertical  ; 
find  the  latitude  of  the  place. 

33.  If  the  difference  between  the  lengths  of  the  longest 
and  shortest  day  at  a  given  place  be  six  hours,  find  the 
latitude. 

34.  If  the  radius  of  the  earth  be  4000  miles,  what  is  the 
area  of  a  spherical  triangle  whose  spherical  excess  is  1°  ? 

35.  If  A",  B",  C"  be  the  chordal  angles  of  the  polar 
triangle  of  ABC,  prove 

cos  A"  =  sin  \  A  cos  (s  —  a),  etc. 

36.  If  the  area  of  a  spherical  triangle  be  one-fourth  the 
area  of  the  sphere,  show  that  the  bisector  of  a  side  is  the 
supplement  of  half  that  side. 

37.  If  the  area  of  a  spherical  triangle  be  one-fourth  the 
area  of  the  sphere,  show  that  the  arcs  joining  the  middle 
points  of  its  sides  are  quadrants. 

38.  Given  the  base  and  area  ;  show  that  the  arc  joining 
the  middle  points  of  the  sides  is  constant  ;  and  if  it  is  a 
quadrant,  then  the  area  of  the  triangle  is  Trr2. 

39.  Two    circles    of    angular   radii,    «  and   /?,    intersect 
orthogonally  on  a  sphere  of  radius  ?*;  find  in  any  manner 
the  area  common  to  the  two. 

40.  If  E  be  the  spherical  excess  of  a  triangle,  prove  that 

etc. 


41.    Show  that  the   sum  of  the  three  arcs   joining  the 
middle  points  of  the  sides  of  the  colunars  is  equal  to  two 


366  SPHERICAL    TEIGONOMETEY. 

right  angles,  the  sides  of  the  original  triangle  being  regarded 
as  the  bases  of  the  colimars. 

42.  Prove  that 

cos2  Ja  sin2S  +  sin2 1  6  sin2(S  -  C)  +  sin2 -Jo  sin2'(S  -  B) 
-f-  2 cos  J a  sinifr  sin  Jc  sinS  sin(S  —  B)  sin(S  —  C)=  1. 

43.  Having  given  the  base  and  the  arc  joining  the  middle 
points  of  the  colunar  on  the  base,  the  circumcircle  is  fixed. 

44.  Prove  sin  \  b  sin  \  c  sin  ( S  —  A)  -f  cos  1  b  cos  J  c  sin  S 

=  cos -Jo. 

45.  If  A  +  B  +  C  =  2  TT,  prove  that 

cos2  -J  a  +  cos2 1  b  +  cos2  J  c  =  1, 
and  cos  C  =  —  cot  -|  a  cot  J  6. 

46.  Solve  the  equations, 

sin  b  cos  c  sin  Z  +  sin  c  cos  6  sin  Y  =  sin  a, 
sin  c  cos  a  sin  X  -f  sin  a  cos  c  sin  Z  '=  sin  6, 
sin  a  cos  6  sin  Y  +  sin  b  cos  a  sin  X  =  sin  c, 
for  sin  X,  sin  Y,  and  sin  Z. 

47.  If  b  and  c  are  constant,  prove  the  following  relations 
between  the  small  variations  of  any  two  parts  of  the  other 
elements  of  the  spherical  triangle  ABC  : 

tfB      tanB  da 


dC      tanC' 

—  =  sinBsinc; 

dA. 

da 

dB 
dA. 

sin  A 

dB 
dA 

sinB  cosC 
sin  A 

dC 

dC 

cosB  sinC 

48.    If  A  and  c  remain  constant,  prove  the  following  : 

^  =  cos  C. 


dB      tanC  db 


EXAMPLES.  367 

49.    If  B  and  C  remain  constant,  prove  the  following  : 

** 


db  db      sin  b  cos  c 

50.    If  A  and  a  remain  constant,  prove  the  following  : 
db  _  tan  b  t  dc  _  tan  c  . 

dB  ~  tan  B  '  dC  ~  tan  C  ' 

db          cosB  d&  sin  b 


dc  cos  C  dC          tan  B  cos  c 

51.  Two  equal  small  circles  are  drawn   touching   each 
other;  show  that  the  angle  between  their  planes  is  twice 
the  complement  of  their  spherical  radius. 

52.  On  a  sphere  whose  radius  is  r  a  small  circle  of  spher- 
ical radius  0  is  described,  and  a  great  circle  is  described 
having  its  pole  on  the  small  circle  ;  show  that  the  length 

o  „          _ 

of  their  common  chord  is  —  -  V  —  cos  2  0. 


53.  Given  the  base  c  of  a  triangle,  and  that 

tan  |-a  tan  \  b  =  tan2^-  B, 
B  being  the  bisector  of  the  base,  find  a  —  b  in  terms  of  c. 

54.  If  C  =  A  +  B,  show  that 

1  —  cos  a  —  cos  b  +  cos  c  =  0. 

55.  If  A  denote  one  of  the  angles  of  an  equilateral  tri 
angle,  and  A'  an  angle  of  its  polar  triangle,  show  that 

cos  A  cos  A'  =  cos  A  +  cos  Af. 

56.  Show  that 

cos  a  cos  B  —  cos  b  cos  A  _  cos  C  +  cos  c 
sin  a  —  sin  b  sin  c 

57.  Prove     cos  A  =  cos  a  sin  b  ~  sin  a  cos  b  cos  C, 

sine 

and      cosA  +  coSB  =  ^!lL(i±*l 

sine 


368  SPHERICAL    TRIGONOMETRY. 

58.  Prove  Legendre's  Theorem  by  means  of  the  relations 

sin  A  _  sin  B  _  sin  C 
sin  a       sin  b       sin  c 

59.  Two  places  are  situated  on  the  same  parallel  of  lati- 
tude </> ;  find  the  difference  of  the  distances  sailed  over  by 
two  ships  passing  between  them,  one  keeping  to  the  great 
circle   course,  the  other  to  the  parallel ;   the  difference  of 
longitude  of  the  places  being  2  A. 

Ans.  2  r  [A  cos  <f>  —  sin"1  (cos  </>  sin  A)  ] . 

60.  If  the  sides  of  a  triangle  be  each  60°,  show  that  the 
circles  described,  each  having  a  vertex  for  pole,  and  passing 
through  the  middle  points  of  the  sides  which  meet  at  it, 
have  the  sides  of   the  supplemental   triangle  for  common 
tangents. 

61.  Find    the  volume  and  also  the  inclination   of   two 
adjacent  faces  (1)  of  a  regular  tetraedron,  (2)  of  a  regular 
octaedron,  (3)  of  a  regular  dodecaedron,  and  (4)  of  a  regular 
icosa'edron,  the  edge  being  one  inch. 

Ans.   (1)  117.85  cu.  in.,    70031'43".4; 

(2)  .4714  cu.  in.,  109°  28' 16"; 

(3)  7.663  cu.  in.,  116°  33' 54"; 

(4)  2.1817  cu.  in.,  138°  11'22".6. 

62.  In  the  tetraedron,  prove   (1)   that  the  circumradius 
is  equal  to  three  times  its  in-radius,  and  (2)  that  the  radius 
of  the  sphere  touching  its  -six  edges  is  a  mean  proportional 
between  the  in-radius  and  circumradius. 

63.  Prove  that  the  ratio  of  the  in-radius  to  the  circum- 
radius is  the  same  in  the  cube  and  the  octaedron,  and  also 
in  the  dodecaedron  and  icosaedron. 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 


This  book  is  due  on  the  last  date  stamped  below, 
or  on  the  date  to  which  renewed.  Renewals  only: 
_,.                                            Tel.  No.  642-3405 
1                    Renewals  may  be  made  4  days  prior  to  date  due. 
Renewed  jbo^ks^are^  subject  to  immediate  recall. 

i 

j^1-^  &  &  ran  7  7 

J6 


W3TDLD 


.-, 


KG'/ 


..« 

160 


Y* 


LD 


General  Library 


911383T 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


